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 a[100001];
int atLast[100001];
int cnt[100001][450];
int conv[100001];
int reconv[450];
int main() {
  int n, m;
  memset(atLast, 0, sizeof(atLast));
  memset(conv, 0, sizeof(conv));
  memset(reconv, 0, sizeof(reconv));
  memset(cnt, 0, sizeof(cnt));
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    if (a[i] <= n) {
      ++atLast[a[i]];
    }
  }
  int legal = 0;
  for (int i = 1; i <= n; ++i) {
    if (atLast[i] >= i) {
      ++legal;
      conv[i] = legal;
      reconv[legal] = i;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= legal; ++j) {
      cnt[i][j] = cnt[i - 1][j];
    }
    if (a[i] <= n && conv[a[i]]) {
      ++cnt[i][conv[a[i]]];
    }
  }
  for (int i = 0; i < m; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    int ret = 0;
    for (int j = 1; j <= legal; ++j) {
      if (cnt[r][j] - cnt[l - 1][j] == reconv[j]) {
        ++ret;
      }
    }
    printf("%d\n", ret);
  }
  return 0;
}
