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>
int n, m, b[105000], a[105000], part[500][105000], c[500], k;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &b[i]);
    if (b[i] > n) continue;
    a[b[i]]++;
  };
  for (int i = 1; i <= n; i++)
    if (a[i] >= i) c[k++] = i;
  for (int i = 0; i < k; i++)
    for (int j = 1; j <= n; j++) part[i][j] = part[i][j - 1] + (b[j] == c[i]);
  for (int i = 0; i < m; i++) {
    int l, r, answer = 0;
    scanf("%d%d", &l, &r);
    for (int j = 0; j < k; j++)
      if (part[j][r] - part[j][l - 1] == c[j]) ++answer;
    printf("%d\n", answer);
  };
};
