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>
struct {
  inline operator int() {
    int x;
    return scanf("%d", &x), x;
  }
} read;
const int maxn = 100005, maxb = 500;
int a[maxn], tot[maxn];
int t[maxb][maxn], val[maxb];
int main() {
  int n = read, q = read;
  for (int i = 1; i <= n; i++)
    if ((a[i] = read) <= n) ++tot[a[i]];
  int p = 0;
  for (int x = 1; x <= n; x++)
    if (tot[x] >= x) {
      val[++p] = x;
      for (int i = 1; i <= n; i++) t[p][i] = t[p][i - 1] + (a[i] == x);
    }
  while (q--) {
    int l = read, r = read, ans = 0;
    for (int i = 1; i <= p; i++)
      if (t[i][r] - t[i][l - 1] == val[i]) ++ans;
    printf("%d\n", ans);
  }
}
