There are n cards (n is even) in the deck. Each card has a positive integer written on it. n / 2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player.

Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible.

The first line of the input contains integer n (2 ≤ n ≤ 100) — the number of cards in the deck. It is guaranteed that n is even.

The second line contains the sequence of n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is equal to the number written on the i-th card.

Print n / 2 pairs of integers, the i-th pair denote the cards that should be given to the i-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input.

It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them.

In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.

In the second sample, all values ai are equal. Thus, any distribution is acceptable.

#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1<<30
#define MOD 1000000007
#define ll long long
#define lson l,m,rt<<1
#define key_value ch[ch[root][1]][0]
#define rson m+1,r,rt<<1|1
#define pi acos(-1.0)
using namespace std;
const int MAXN = 1010;
int a[MAXN],n,ans[MAXN],vis[MAXN];
vector<int>s;
int main(){while(~scanf("%d",&n)){int sum = 0;for(int i = 1; i <= n; i++){scanf("%d",&a[i]);sum += a[i];}s.clear();sum /= (n / 2);memset(vis,0,sizeof(vis));for(int i = 1; i <= n; i++){if(vis[i])continue;vis[i] = 1;for(int j = 1; j <= n; j++){if(!vis[j] && a[i] + a[j] == sum){ans[i] = j;vis[j] = vis[i] = 1;s.push_back(i);break;}}}for(int i = 0; i < s.size(); i++){printf("%d %d\n",s[i],ans[s[i]]);}}return 0;
}

Vasya has the square chessboard of size n × n and m rooks. Initially the chessboard is empty. Vasya will consequently put the rooks on the board one after another.

The cell of the field is under rook's attack, if there is at least one rook located in the same row or in the same column with this cell. If there is a rook located in the cell, this cell is also under attack.

You are given the positions of the board where Vasya will put rooks. For each rook you have to determine the number of cells which arenot under attack after Vasya puts it on the board.

The first line of the input contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ min(100 000, n2)) — the size of the board and the number of rooks.

Each of the next m lines contains integers xi and yi (1 ≤ xi, yi ≤ n) — the number of the row and the number of the column where Vasya will put the i-th rook. Vasya puts rooks on the board in the order they appear in the input. It is guaranteed that any cell will contain no more than one rook.

Print m integer, the i-th of them should be equal to the number of cells that are not under attack after first i rooks are put.

#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1<<30
#define MOD 1000000007
#define ll long long
#define lson l,m,rt<<1
#define key_value ch[ch[root][1]][0]
#define rson m+1,r,rt<<1|1
#define pi acos(-1.0)
using namespace std;
const int MAXN = 100010;
int vis[MAXN][2],n,m;
int main(){while(~scanf("%d%d",&n,&m)){memset(vis,0,sizeof(vis));ll ans = 0;int rownum = 0,colnum = 0;for(int i = 1; i <= m; i++){int x,y,flag1 = 0,flag2 = 0;scanf("%d%d",&x,&y);if(!vis[y][1]){vis[y][1] = 1; ans += n; flag1 ++;ans -= rownum;if(vis[x][0])ans += 1;}if(!vis[x][0]){vis[x][0] = 1; ans += n; flag2 ++;ans -= colnum;if(!flag1)ans += 1;}if(vis[x][0] && vis[y][1] && (flag1 || flag2)){ans -= 1;}if(flag1)colnum ++;if(flag2)rownum ++;//cout<<ans<<' '<<endl;printf("%lld ",1LL*n*n - ans);}puts("");}return 0;
}

Sergei B., the young coach of Pokemons, has found the big house which consists of n flats ordered in a row from left to right. It is possible to enter each flat from the street. It is possible to go out from each flat. Also, each flat is connected with the flat to the left and the flat to the right. Flat number 1 is only connected with the flat number 2 and the flat number n is only connected with the flat numbern - 1.

There is exactly one Pokemon of some type in each of these flats. Sergei B. asked residents of the house to let him enter their flats in order to catch Pokemons. After consulting the residents of the house decided to let Sergei B. enter one flat from the street, visit several flats and then go out from some flat. But they won't let him visit the same flat more than once.

Sergei B. was very pleased, and now he wants to visit as few flats as possible in order to collect Pokemons of all types that appear in this house. Your task is to help him and determine this minimum number of flats he has to visit.

The first line contains the integer n (1 ≤ n ≤ 100 000) — the number of flats in the house.

The second line contains the row s with the length n, it consists of uppercase and lowercase letters of English alphabet, the i-th letter equals the type of Pokemon, which is in the flat number i.

Print the minimum number of flats which Sergei B. should visit in order to catch Pokemons of all types which there are in the house.

#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1<<30
#define MOD 1000000007
#define ll long long
#define lson l,m,rt<<1
#define key_value ch[ch[root][1]][0]
#define rson m+1,r,rt<<1|1
#define pi acos(-1.0)
using namespace std;
const int MAXN = 100010;
char s[MAXN];
int n,vis[200];
int main(){while(~scanf("%d",&n)){int all = 0;scanf("%s",s+1);memset(vis,0,sizeof(vis));for(int i = 1; i <= n; i++){if(!vis[s[i]]){all += 1, vis[s[i]] = 1;}}memset(vis,0,sizeof(vis));int num = 0;int k,j;int minlen = INF,flag = 0;k = 1;j = 1;for(; j <= n;){if(!flag){if(!vis[s[j]]){num ++;}vis[s[j]] ++;if(num == all){minlen = min(minlen,j - k + 1);flag = 1;continue;}j ++;} else {while(k <= j){if(num == all)minlen = min(minlen,j - k + 1);if(vis[s[k]] - 1 > 0){vis[s[k]] --;k ++;} else {vis[s[k]] = 0;k ++;num -= 1;flag = 0; j += 1; break;}}}}if(j > n)j = n;//cout<<j<<' '<<k<<' '<<num<<endl;while(k <= j){if(num == all)minlen = min(minlen,j - k + 1);if(vis[s[k]] - 1 > 0){vis[s[k]] --;k ++;} else {break;}}printf("%d\n",minlen);}return 0;
}

Treeland is a country in which there are n towns connected by n - 1 two-way road such that it's possible to get from any town to any other town.

In Treeland there are 2k universities which are located in different towns.

Recently, the president signed the decree to connect universities by high-speed network.The Ministry of Education understood the decree in its own way and decided that it was enough to connect each university with another one by using a cable. Formally, the decree will be done!

To have the maximum sum in the budget, the Ministry decided to divide universities into pairs so that the total length of the required cable will be maximum. In other words, the total distance between universities in k pairs should be as large as possible.

Help the Ministry to find the maximum total distance. Of course, each university should be present in only one pair. Consider that all roads have the same length which is equal to 1.

The first line of the input contains two integers n and k (2 ≤ n ≤ 200 000, 1 ≤ k ≤ n / 2) — the number of towns in Treeland and the number of university pairs. Consider that towns are numbered from 1 to n.

The second line contains 2k distinct integers u1, u2, ..., u2k (1 ≤ ui ≤ n) — indices of towns in which universities are located.

The next n - 1 line contains the description of roads. Each line contains the pair of integers xj and yj (1 ≤ xj, yj ≤ n), which means that the j-th road connects towns xj and yj. All of them are two-way roads. You can move from any town to any other using only these roads.

Print the maximum possible sum of distances in the division of universities into k pairs.

The figure below shows one of possible division into pairs in the first test. If you connect universities number 1 and 6 (marked in red) and universities number 2 and 5 (marked in blue) by using the cable, the total distance will equal 6 which will be the maximum sum in this example.

#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1<<30
#define MOD 1000000007
#define ll long long
#define lson l,m,rt<<1
#define key_value ch[ch[root][1]][0]
#define rson m+1,r,rt<<1|1
#define pi acos(-1.0)
using namespace std;
const int MAXN = 200010;
struct node{int to;int next;
}edge[MAXN*2];
int ind,pre[MAXN],vis[MAXN];
ll ans,num[MAXN];
int n,k;
void add(int x,int y){edge[ind].to = y;edge[ind].next = pre[x];pre[x] = ind ++;
}
void dfs(int rt){vis[rt] = 1;for(int i = pre[rt]; i != -1; i = edge[i].next){int t = edge[i].to;if(!vis[t]){dfs(t);num[rt] += num[t];ans += min(num[t],2*k-num[t]);}}
}
int main(){while(~scanf("%d%d",&n,&k)){memset(num,0,sizeof(num));for(int i = 1; i <= 2 * k; i++){int x;scanf("%d",&x);num[x] = 1;}ind = 0;memset(pre,-1,sizeof(pre));for(int i = 1; i < n; i++){int x,y;scanf("%d%d",&x,&y);add(x,y),add(y,x);}ans = 0;memset(vis,0,sizeof(vis));dfs(1);printf("%lld\n",ans);}return 0;
}