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 = 111111;
int sum[500][maxn], a[maxn], b[maxn], c[500];
int main() {
  int n, m;
  while (scanf("%d%d", &n, &m) != EOF) {
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      if (a[i] <= n) b[a[i]]++;
    }
    int t = 0, num = 0;
    for (int i = 1; i <= n; i++) {
      if (b[i] >= i) c[t++] = i;
    }
    for (int i = 0; i < t; i++) {
      for (int j = 1; j <= n; j++)
        if (c[i] == a[j - 1])
          sum[i][j] = sum[i][j - 1] + 1;
        else
          sum[i][j] = sum[i][j - 1];
    }
    while (m--) {
      int l, r;
      scanf("%d%d", &l, &r);
      num = 0;
      for (int i = 0; i < t; i++)
        if (sum[i][r] - sum[i][l - 1] == c[i]) num++;
      printf("%d\n", num);
    }
  }
  return 0;
}
