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, sqr;
int a[100005];
int cnt[100005];
int res[100005];
int ans = 0;
struct que {
  int l, r, idx;
} q[100005];
bool operator<(const que &q1, const que &q2) {
  if (q1.l / sqr != q2.l / sqr) return q1.l < q2.l;
  if ((q1.l / sqr) % 2) return q1.r > q2.r;
  return q1.r < q2.r;
}
int main() {
  memset(res, 0, sizeof(res));
  scanf("%d %d", &n, &m);
  sqr = sqrt(m);
  if (sqr * sqr != m) sqr++;
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &q[i].l, &q[i].r);
    q[i].l;
    q[i].r;
    q[i].idx = i;
  }
  sort(q, q + m);
  int l = 1, r = 0;
  for (int i = 0; i < m; i++) {
    while (q[i].r > r) {
      r++;
      if (a[r] <= 100000) {
        if (cnt[a[r]] == a[r]) ans--;
        cnt[a[r]]++;
        if (cnt[a[r]] == a[r]) ans++;
      }
    }
    while (q[i].l < l) {
      l--;
      if (a[l] <= 100000) {
        if (cnt[a[l]] == a[l]) ans--;
        cnt[a[l]]++;
        if (cnt[a[l]] == a[l]) ans++;
      }
    }
    while (q[i].r < r) {
      if (a[r] <= 100000) {
        if (cnt[a[r]] == a[r]) ans--;
        cnt[a[r]]--;
        if (cnt[a[r]] == a[r]) ans++;
      }
      r--;
    }
    while (q[i].l > l) {
      if (a[l] <= 100000) {
        if (cnt[a[l]] == a[l]) ans--;
        cnt[a[l]]--;
        if (cnt[a[l]] == a[l]) ans++;
      }
      l++;
    }
    res[q[i].idx] = ans;
  }
  for (int i = 0; i < m; i++) printf("%d\n", res[i]);
  return 0;
}
