The Little Elephant loves playing with arrays. He has array a, consisting of n positive integers, indexed from 1 to n. Let's denote the number with index i as ai. 

Additionally the Little Elephant has m queries to the array, each query is characterised by a pair of integers lj and rj (1 ≤ lj ≤ rj ≤ n). For each query lj, rj the Little Elephant has to count, how many numbers x exist, such that number x occurs exactly x times among numbers alj, alj + 1, ..., arj.

Help the Little Elephant to count the answers to all queries.

Input

The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the size of array a and the number of queries to it. The next line contains n space-separated positive integers a1, a2, ..., an (1 ≤ ai ≤ 109). Next m lines contain descriptions of queries, one per line. The j-th of these lines contains the description of the j-th query as two space-separated integers lj and rj (1 ≤ lj ≤ rj ≤ n).

Output

In m lines print m integers — the answers to the queries. The j-th line should contain the answer to the j-th query.

Examples

Input

7 2
3 1 2 2 3 3 7
1 7
3 4


Output

3
1
#include <bits/stdc++.h>
using namespace std;
int cnt[100005];
int gg[500];
int gn;
int ww[100005];
int cc[100005][455];
int n, m;
int dd[100005];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &dd[i]);
    if (dd[i] <= n) {
      cnt[dd[i]]++;
      if (cnt[dd[i]] >= dd[i] && !ww[dd[i]]) {
        gg[++gn] = dd[i];
        ww[dd[i]] = gn;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    memcpy(cc[i], cc[i - 1], sizeof(cc[i - 1]));
    if (dd[i] <= n && ww[dd[i]]) {
      cc[i][ww[dd[i]]]++;
    }
  }
  int a, b;
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &a, &b);
    int res = 0;
    for (int j = 1; j <= gn; ++j) {
      if (cc[b][j] - cc[a - 1][j] == gg[j]) res++;
    }
    printf("%d\n", res);
  }
  return 0;
}
