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;
int s[500];
int d[500][100010];
int a[100100];
map<int, int> p;
map<int, int> pa;
map<int, int> paa;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  int i;
  int j = 1;
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    p[a[i]]++;
    if (p[a[i]] == a[i]) {
      s[j] = a[i];
      pa[a[i]] = j;
      d[j][i] = a[i];
      j++;
    } else if (pa[a[i]] > 0) {
      d[pa[a[i]]][i] = p[a[i]];
    }
  }
  int l, r;
  int H = j;
  int ans = 0;
  for (i = 1; i <= n; i++) {
    if (pa[a[i]] > 0 && p[a[i]] > paa[a[i]]) {
      d[pa[a[i]]][i] = paa[a[i]] + 1;
      paa[a[i]]++;
    }
  }
  for (i = 1; i < H; i++) {
    for (j = 1; j <= n; j++) {
      if (d[i][j] < d[i][j - 1]) d[i][j] = d[i][j - 1];
    }
  }
  for (i = 0; i < m; i++) {
    scanf("%d%d", &l, &r);
    ans = 0;
    for (j = 1; j < H; j++) {
      if (d[j][r] - d[j][l - 1] == s[j]) ans++;
    }
    printf("%d\n", ans);
  }
  return 0;
}
