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 N = 1e5 + 5;
int n, q, res, block, a[N], ans[N], vis[N];
struct node {
  int l, r, id;
  bool operator<(const node& o) const {
    if (l / block == o.l / block) return r < o.r;
    return l / block < o.l / block;
  }
} p[N];
void add(int x) {
  if (x == 0) return;
  if (a[x] > n) return;
  vis[a[x]]++;
  if (vis[a[x]] == a[x])
    res++;
  else if (vis[a[x]] == a[x] + 1)
    res--;
}
void del(int x) {
  if (x == 0) return;
  if (a[x] > n) return;
  vis[a[x]]--;
  if (vis[a[x]] == a[x]) res++;
  if (vis[a[x]] == a[x] - 1) res--;
}
int main() {
  int t = 1;
  for (int cas = 1; cas <= t; cas++) {
    memset(vis, 0, sizeof(vis));
    res = 0;
    scanf("%d%d", &n, &q);
    block = sqrt(n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= q; i++) scanf("%d%d", &p[i].l, &p[i].r), p[i].id = i;
    sort(p + 1, p + 1 + q);
    int l = 0, r = 0;
    for (int i = 1; i <= q; i++) {
      while (l < p[i].l) del(l++);
      while (l > p[i].l) add(--l);
      while (r > p[i].r) del(r--);
      while (r < p[i].r) add(++r);
      ans[p[i].id] = res;
    }
    for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  }
}
