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[100005], b[100005], c[100005], num[100005];
int s;
struct Query {
  int l, r, id;
  bool operator<(const Query &a) const {
    if (l / s == a.l / s) return r < a.r;
    return l / s < a.l / s;
  }
} q[100005];
int ans[100005];
int main() {
  int n, m;
  while (scanf("%d%d", &n, &m) == 2) {
    s = (int)sqrt(1.0 * n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
    sort(b + 1, b + n + 1);
    for (int i = 1; i <= n; i++) c[i] = lower_bound(b + 1, b + 1 + n, a[i]) - b;
    for (int i = 1; i <= m; i++) scanf("%d%d", &q[i].l, &q[i].r), q[i].id = i;
    sort(q + 1, q + m + 1);
    int l = 1, r = 0, tmp = 0;
    for (int i = 1; i <= m; i++) {
      while (r < q[i].r) {
        r++;
        if (num[c[r]] == a[r] - 1)
          tmp++;
        else if (num[c[r]] == a[r])
          tmp--;
        num[c[r]]++;
      }
      while (r > q[i].r) {
        if (num[c[r]] == a[r])
          tmp--;
        else if (num[c[r]] == a[r] + 1)
          tmp++;
        num[c[r]]--;
        r--;
      }
      while (l < q[i].l) {
        if (num[c[l]] == a[l])
          tmp--;
        else if (num[c[l]] == a[l] + 1)
          tmp++;
        num[c[l]]--;
        l++;
      }
      while (l > q[i].l) {
        l--;
        if (num[c[l]] == a[l] - 1)
          tmp++;
        else if (num[c[l]] == a[l])
          tmp--;
        num[c[l]]++;
      }
      ans[q[i].id] = tmp;
    }
    for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  }
  return 0;
}
