Mr. Frog has an integer sequence of length n, which can be denoted as a1,a2,⋯,an There are m queries.

In the i-th query, you are given two integers li and ri. Consider the subsequence ali,ali+1,ali+2,⋯,ari.

We can denote the positions(the positions according to the original sequence) where an integer appears first in this subsequence as p(i)1,p(i)2,⋯,p(i)ki (in ascending order, i.e.,p(i)1<p(i)2<⋯<p(i)ki).

Note that ki is the number of different integers in this subsequence. You should output p(i)⌈ki2⌉for the i-th query.

In the first line of input, there is an integer T (T≤2) denoting the number of test cases.

Each test case starts with two integers n (n≤2×105) and m (m≤2×105). There are n integers in the next line, which indicate the integers in the sequence(i.e., a1,a2,⋯,an,0≤ai≤2×105).

There are two integers li and ri in the following m lines.

However, Mr. Frog thought that this problem was too young too simple so he became angry. He modified each query to l‘i,r‘i(1≤l‘i≤n,1≤r‘i≤n). As a result, the problem became more exciting.

We can denote the answers as ans1,ans2,⋯,ansm. Note that for each test case ans0=0.

You can get the correct input li,ri from what you read (we denote them as l‘i,r‘i)by the following formula:

You should output one single line for each test case.

For each test case, output one line “Case #x: p1,p2,⋯,pm”, where x is the case number (starting from 1) and p1,p2,⋯,pm is the answer.

2
5 2
3 3 1 5 4
2 2
4 4
5 2
2 5 2 1 2
2 3
2 4

Case #1: 3 3
Case #2: 3 1 

题意：给一个数列，有m个询问，每次问[l,r]区间中所有数的第一次出现的位置的中位数是多少（有点绕），强制在线。

分析：比赛时队友黄大爷临时想出了一个分块的做法卡过了这题Orz，这题的一般做法是在主席树上做二分，倒着插点建树，然后对于每个询问的l，我们都可以得到一个从l开始的权值线段树，表示每个位置上的数是否是第一次出现，然后我们对于每个询问l,r，就可以二分它的中位数的位置了 直接在权值线段树中递归下去就可以了。

#include <cstdio>
#include <iostream>
#include <algorithm>3e
#include <cstdlib>
#include <cstdio>
#include <set>
#include <map>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <unordered_map>
#include <cstring>
#define MOD 1000000007
#define N 200005
using namespace std;
typedef long long ll;
int T,n,m,cnt,pos,a[N],b[N],rt[N],f[N];
struct Tree
{int ls,rs,sum;
}tr[N*40];
void Build(int &node,int l,int r)
{node = ++cnt;tr[node].sum = 0;if(l == r) return;int mid = (l+r)>>1;Build(tr[node].ls,l,mid);Build(tr[node].rs,mid+1,r);
}
void Insert(int pre,int &node,int x,int l,int r)
{node = ++cnt;tr[node] = tr[pre];tr[node].sum += x;if(l == r) return;int mid = (l + r) >> 1;if(pos <= mid) Insert(tr[pre].ls,tr[node].ls,x,l,mid);else Insert(tr[pre].rs,tr[node].rs,x,mid+1,r);
}
int Ask(int node,int x,int l,int r)
{if(l == r) return tr[node].sum;int mid = (l+r)>>1;if(x <= mid) return Ask(tr[node].ls,x,l,mid);else return tr[tr[node].ls].sum + Ask(tr[node].rs,x,mid+1,r);
}
int Query(int node,int k,int l,int r)
{if(l == r) return l;int mid = (l+r)>>1;if(tr[tr[node].ls].sum >= k) return Query(tr[node].ls,k,l,mid);else return Query(tr[node].rs,k-tr[tr[node].ls].sum,mid+1,r);
}
int main()
{scanf("%d",&T);for(int t = 1;t <= T;t++){cnt = 0;memset(f,0,sizeof(f));scanf("%d%d",&n,&m);for(int i = 1;i <= n;i++) scanf("%d",&a[i]);printf("Case #%d:",t);Build(rt[n+1],1,n);for(int i = n;i;i--){pos = f[a[i]];if(pos){Insert(rt[i+1],rt[i],-1,1,n);pos = i;Insert(rt[i],rt[i],1,1,n);}else{pos = i;Insert(rt[i+1],rt[i],1,1,n);}f[a[i]] = i;}int pre = 0;for(int i = 1;i <= m;i++){int l,r;scanf("%d%d",&l,&r);l = (l + pre) % n + 1,r = (r + pre) % n + 1;if(l > r) swap(l,r);int k = Ask(rt[l],r,1,n);pre = Query(rt[l],(k+1)/2,1,n);printf(" %d",pre);}printf("\n");}
}