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 = 100004;
int a[N];
struct node {
  int l, r, id;
} qry[N];
int block;
bool compare(node a, node b) {
  if ((a.l / block) != (b.l / block))
    return a.l < b.l;
  else
    return a.r < b.r;
}
int cnt;
int ans[N], mp[N];
void update(int i, int val) {
  if (mp[a[i]] == a[i]) cnt--;
  mp[a[i]] += val;
  if (mp[a[i]] == a[i]) cnt++;
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  int i;
  block = sqrt(n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (i = 1; i <= m; i++) {
    int left, right;
    scanf("%d %d", &left, &right);
    qry[i].l = left;
    qry[i].r = right;
    qry[i].id = i;
  }
  sort(qry + 1, qry + m + 1, compare);
  int left = qry[1].l;
  int right = qry[1].l - 1;
  for (i = 1; i <= m; i++) {
    while (left < qry[i].l) {
      if (a[left] < N) update(left, -1);
      left++;
    }
    while (left > qry[i].l) {
      left--;
      if (a[left] < N) update(left, 1);
    }
    while (right < qry[i].r) {
      right++;
      if (a[right] < N) update(right, 1);
    }
    while (right > qry[i].r) {
      if (a[right] < N) update(right, -1);
      right--;
    }
    ans[qry[i].id] = cnt;
  }
  for (i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}
