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;
const int MAXN = 100005;
int a[MAXN];
int sum[MAXN];
vector<int> has[MAXN];
int sz[MAXN];
int n, m;
void add(int p, int v) {
  while (p <= n) {
    sum[p] += v;
    p += ((-p) & p);
  }
}
int call(int p) {
  int res = 0;
  while (p >= 1) {
    res += sum[p];
    p -= ((-p) & p);
  }
  return res;
}
struct node {
  int l, r, id, ans;
} q[MAXN];
bool cmp(node a, node b) { return a.r < b.r; }
bool cmp2(node a, node b) { return a.id < b.id; }
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  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, cmp);
  int p = 1;
  for (int i = 1; i <= m; i++) {
    while (p <= q[i].r) {
      if (a[p] <= n) {
        has[a[p]].push_back(p);
        sz[a[p]]++;
        if (sz[a[p]] >= a[p]) {
          add(has[a[p]][sz[a[p]] - a[p]], 1);
        }
        if (sz[a[p]] >= a[p] + 1) {
          add(has[a[p]][sz[a[p]] - a[p] - 1], -2);
        }
        if (sz[a[p]] >= a[p] + 2) {
          add(has[a[p]][sz[a[p]] - a[p] - 2], 1);
        }
      }
      p++;
    }
    q[i].ans = call(q[i].r) - call(q[i].l - 1);
  }
  sort(q + 1, q + m + 1, cmp2);
  for (int i = 1; i <= m; i++) {
    printf("%d\n", q[i].ans);
  }
  return 0;
}
