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;
struct node {
  int x, num;
} a[100005];
int n, t, m, k, i, j, s[100005], b[450], h[100005][450];
map<int, int> mymap;
int main() {
  while (~scanf("%d%d", &n, &t)) {
    mymap.clear();
    m = 0;
    for (i = 1; i <= n; i++) scanf("%d", &s[i]);
    for (i = 1; i <= n; i++) {
      if (mymap.find(s[i]) == mymap.end()) {
        mymap[s[i]] = ++m;
        a[m].x = s[i];
        a[m].num = 0;
      }
      a[mymap[s[i]]].num++;
    }
    k = 0;
    for (i = 1; i <= m; i++)
      if (a[i].x <= a[i].num) b[++k] = a[i].x;
    for (i = 1; i <= k; i++) {
      h[0][i] = 0;
      for (j = 1; j <= n; j++)
        if (s[j] == b[i])
          h[j][i] = h[j - 1][i] + 1;
        else
          h[j][i] = h[j - 1][i];
    }
    int l, r, ans;
    while (t--) {
      ans = 0;
      scanf("%d%d", &l, &r);
      for (i = 1; i <= k; i++)
        if (h[r][i] - h[l - 1][i] == b[i]) ans++;
      printf("%d\n", ans);
    }
  }
  return 0;
}
