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 n, m;
int arr[100005];
int rep[100005];
int ansr[100005];
struct query {
  int l, r, idx;
};
int block;
bool comp(query x, query y) {
  if (x.l / block != y.l / block) return x.l / block < y.l / block;
  return x.r < y.r;
}
int main() {
  cin >> n >> m;
  block = sqrt(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    if (arr[i] > 1e5) arr[i] = -1;
  }
  vector<query> B(m);
  for (int i = 0; i < m; i++)
    scanf("%d %d", &B[i].l, &B[i].r), B[i].idx = i, B[i].l--, B[i].r--;
  sort(B.begin(), B.end(), comp);
  int ans = 0;
  int currL = 0;
  int currR = 0;
  for (int i = 0; i < m; i++) {
    int L = B[i].l, R = B[i].r;
    while (currL < L) {
      rep[arr[currL]]--;
      if (arr[currL] != -1) {
        if (rep[arr[currL]] == arr[currL])
          ans++;
        else if (rep[arr[currL]] == arr[currL] - 1)
          ans--;
      }
      currL++;
    }
    while (currL > L) {
      rep[arr[currL - 1]]++;
      if (arr[currL] != -1) {
        if (rep[arr[currL - 1]] == arr[currL - 1])
          ans++;
        else if (rep[arr[currL - 1]] == arr[currL - 1] + 1)
          ans--;
      }
      currL--;
    }
    while (currR <= R) {
      rep[arr[currR]]++;
      if (arr[currL] != -1) {
        if (rep[arr[currR]] == arr[currR])
          ans++;
        else if (rep[arr[currR]] == arr[currR] + 1)
          ans--;
      }
      currR++;
    }
    while (currR > R + 1) {
      rep[arr[currR - 1]]--;
      if (arr[currL] != -1) {
        if (rep[arr[currR - 1]] == arr[currR - 1])
          ans++;
        else if (rep[arr[currR - 1]] == arr[currR - 1] - 1)
          ans--;
      }
      currR--;
    }
    ansr[B[i].idx] = ans;
  }
  for (int i = 0; i < m; i++) printf("%d\n", ansr[i]);
  return 0;
}
