Given an array, your task is to find the k-th occurrence (from left to right) of an integer v. To make the problem more difficult (and interesting!), you‘ll have to answer m such queries.

Input

There are several test cases. The first line of each test case contains two integers n, m(1<=n,m<=100,000), the number of elements in the array, and the number of queries. The next line contains n positive integers not larger than 1,000,000. Each of the following m lines contains two integer k and v (1<=k<=n, 1<=v<=1,000,000). The input is terminated by end-of-file (EOF). The size of input file does not exceed 5MB.

Output

For each query, print the 1-based location of the occurrence. If there is no such element, output 0 instead.

Sample Input

8 4
1 3 2 2 4 3 2 1
1 3
2 4
3 2
4 2

Output for the Sample Input

2
0
7
0

题意：给出一个数组，m次询问，问第k个v的位置，如果不存在，输出0.

先预处理排序，然后二分求下界。

#include <string>
#include <iostream>
#include <map>
#include <cstdio>
#include <algorithm>
using namespace std;
int n,m,k,v;
struct C {
    int pos,num;
}a[100005];
bool cmp(C a, C b) {
    if(a.num == b.num) return a.pos < b.pos;
    return a.num < b.num;
}
int bs(int v) {
    int m, x = 1, y = n;
    while(x < y) {
        m = x+(y-x)/2;
        if(a[m].num >= v) y = m;
        else x = m+1;
    }
    return x;
}
int main()
{
    while(~scanf("%d%d",&n,&m)) {
        for(int i=1;i<=n;i++) {
            scanf("%d",&a[i].num);
            a[i].pos = i;
        }
        sort(a+1, a+1+n, cmp);
        for(int i=1; i<=m; i++) {
            scanf("%d%d",&k,&v);
            int p = bs(v);
            if(p+k-1 <= n && a[p+k-1].num == v) printf("%d\n",a[p+k-1].pos);
            else printf("0\n");
        }
    }
    return 0;
}