【CodeForces - 1020C】Elections （枚举投票数，贪心）

题干：

As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.

Elections are coming. You know the number of voters and the number of parties — nnand mm respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give ii-th voter cici bytecoins you can ask him to vote for any other party you choose.

The United Party of Berland has decided to perform a statistical study — you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.

Input

The first line of input contains two integers nn and mm (1≤n,m≤30001≤n,m≤3000) — the number of voters and the number of parties respectively.

Each of the following nn lines contains two integers pipi and cici (1≤pi≤m1≤pi≤m, 1≤ci≤1091≤ci≤109) — the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.

The United Party of Berland has the index 11.

Output

Print a single number — the minimum number of bytecoins needed for The United Party of Berland to win the elections.

Examples

Input

1 2
1 100

Output

Input

Output

Input

Output

Note

In the first sample, The United Party wins the elections even without buying extra votes.

In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 33, 44 and 55 get one vote and party number 22 gets no votes.

In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.

题目大意：

n个人，m个党派，第i个人开始想把票投给党派pi，而如果想让他改变他的想法需要花费ci元。你现在是党派1，问你最少花多少钱使得你的党派得票数大于其它任意党派。

解题报告：

一般思路想到就是二分+check，不难发现这题是没有单调性的，而且时间复杂度允许，我们就从小到大枚举得票数，维护一个最小话费就行了

AC代码：

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int MAXN = 10000 + 5;
vector<ll > a[MAXN];
ll cnt[MAXN],bk[MAXN];
pair<ll , ll> p[MAXN];
int main()
{int n,m;cin>>m>>n;for(int j = 0; j<n; j++) {scanf("%lld%lld",&p[j].second,&p[j].first);p[j].second--;cnt[p[j].second]++;}sort(p,p+n);ll ans = LLONG_MAX;for(int v = 1; v <=n; v++) {memset(cnt,0,sizeof(cnt));memset(bk,0,sizeof(bk));for(int j = 0; j<n; j++) cnt[p[j].second]++;ll chu = cnt[0],cost = 0;for(int i = 0; i<n; i++) {if(p[i].second !=0 && cnt[p[i].second ] >= v) {cost += p[i].first;chu++;cnt[p[i].second]--;bk[i] = 1;}}for(int i = 0; i<n; i++) {if(p[i].second !=0 && chu < v && bk[i] == 0) {chu++;cost += p[i].first;}}ans = min(ans,cost);}printf("%lld\n",ans);return 0 ;
}

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