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;
long long Block;
struct node {
  long long L, R, idx;
} Q[100100];
bool cmp(node x, node y) {
  if (x.L / Block != y.L / Block) {
    return x.L / Block < y.L / Block;
  }
  return x.R < y.R;
}
int main() {
  long long N, m;
  cin >> N >> m;
  long long a[N + 50];
  for (int i = 1; i <= N; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < m; i++) {
    long long l, r;
    cin >> l >> r;
    Q[i].L = l, Q[i].R = r, Q[i].idx = i + 1;
  }
  Block = sqrt(N);
  sort(Q, Q + m, cmp);
  long long cnt[N + 50];
  memset(cnt, 0, sizeof(cnt));
  long long ans = 0;
  long long res[m + 5];
  long long CL = 1, CR = 1;
  if (a[CL] <= N) {
    cnt[a[CL]]++;
    if (cnt[a[CL]] == a[CL]) ans++;
  }
  for (int i = 0; i < m; i++) {
    int L = Q[i].L;
    int R = Q[i].R;
    while (CL < L) {
      if (a[CL] > N)
        int nana = 15;
      else {
        int p = cnt[a[CL]];
        cnt[a[CL]]--;
        if (a[CL] == cnt[a[CL]])
          ans++;
        else if (cnt[a[CL]] < a[CL] and p == a[CL])
          ans--;
      }
      CL++;
    }
    while (CL > L) {
      CL--;
      if (a[CL] > N)
        int nana = 15;
      else {
        int p = cnt[a[CL]];
        cnt[a[CL]]++;
        if (a[CL] == cnt[a[CL]])
          ans++;
        else if (cnt[a[CL]] > a[CL] and p == a[CL])
          ans--;
      }
    }
    while (CR > R) {
      if (a[CR] > N)
        int nana = 15;
      else {
        int p = cnt[a[CR]];
        cnt[a[CR]]--;
        if (a[CR] == cnt[a[CR]])
          ans++;
        else if (cnt[a[CR]] < a[CR] and p == a[CR])
          ans--;
      }
      CR--;
    }
    while (CR < R) {
      CR++;
      if (a[CR] > N)
        int nana = 15;
      else {
        int p = cnt[a[CR]];
        cnt[a[CR]]++;
        if (a[CR] == cnt[a[CR]])
          ans++;
        else if (cnt[a[CR]] > a[CR] and p == a[CR])
          ans--;
      }
    }
    res[Q[i].idx] = ans;
  }
  for (int i = 1; i <= m; i++) cout << res[i] << '\n';
}
