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 = 100010;
int n, m, cnt;
int d[500][maxn];
int f[maxn], cn[maxn];
map<int, int> mp;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &f[i]);
    mp[f[i]]++;
  }
  cnt = 0;
  for (map<int, int>::iterator it = mp.begin(); it != mp.end(); it++)
    if (it->first <= it->second) {
      cnt++;
      cn[cnt] = it->first;
    }
  for (int i = 0; i < n; i++) {
    for (int j = 1; j <= cnt; j++)
      if (cn[j] == f[i])
        d[j][i + 1] = d[j][i] + 1;
      else
        d[j][i + 1] = d[j][i];
  }
  int x, y, ans, i;
  while (m--) {
    ans = 0;
    scanf("%d%d", &x, &y);
    for (i = 1; i <= y - x + 1 && i <= cnt; i++) {
      if (d[i][y] - d[i][x - 1] == cn[i]) ans++;
    }
    printf("%d\n", ans);
  }
  return 0;
}
