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 = 100000;
int n, m;
vector<int> pos[MAXN + 10];
vector<int> T;
int main() {
  int i, j, x, y, cnt, p, t;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (x <= n) pos[x].push_back(i);
  }
  for (i = 1; i <= n; i++)
    if (pos[i].size() >= i) T.push_back(i);
  for (i = 1; i <= m; i++) {
    scanf("%d%d", &x, &y);
    cnt = 0;
    for (j = 0; j < T.size(); j++) {
      p = T[j];
      t = (upper_bound(pos[p].begin(), pos[p].end(), y) - pos[p].begin()) -
          (lower_bound(pos[p].begin(), pos[p].end(), x) - pos[p].begin());
      if (t == p) cnt++;
    }
    printf("%d\n", cnt);
  }
  return 0;
}
