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 dp[500][100005];
int n, q;
int a[100005], b[100005];
int c[100005][2];
int main() {
  while (scanf("%d%d", &n, &q) != EOF) {
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      b[i] = a[i];
    }
    int cnt = -1;
    sort(b + 1, b + 1 + n);
    b[0] = -1;
    memset(c, 0, sizeof(cnt));
    for (int i = 1; i <= n; i++) {
      if (b[i] != b[i - 1]) cnt++;
      c[cnt][0] = b[i];
      c[cnt][1]++;
    }
    vector<int> v;
    for (int i = 0; i <= cnt; i++) {
      if (c[i][1] >= c[i][0]) v.push_back(c[i][0]);
    }
    for (int i = 0; i < v.size(); i++) {
      dp[i][0] = 0;
      for (int j = 1; j <= n; j++) {
        if (a[j] == v[i])
          dp[i][j] = dp[i][j - 1] + 1;
        else
          dp[i][j] = dp[i][j - 1];
      }
    }
    while (q--) {
      int l, r;
      scanf("%d%d", &l, &r);
      int ans = 0;
      for (int i = 0; i < v.size(); i++)
        ans += (dp[i][r] - dp[i][l - 1]) == v[i];
      printf("%d\n", ans);
    }
  }
  return 0;
}
