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 n, m, dp[500][100005], num[100005], vis[100005], st[100005], a[500];
int main() {
  memset(dp, 0, sizeof(0));
  memset(vis, 0, sizeof(vis));
  memset(st, 0, sizeof(st));
  scanf("%d%d", &n, &m);
  int top = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &num[i]);
    if (num[i] <= n) {
      vis[num[i]]++;
      if (vis[num[i]] == num[i]) {
        st[num[i]] = top;
        a[top++] = num[i];
      }
    }
  }
  for (int i = 0; i < n; i++) {
    int v = num[i + 1];
    for (int j = 1; j < top; j++) dp[j][i + 1] = dp[j][i];
    if (v <= n && vis[v] >= v) dp[st[v]][i + 1]++;
  }
  while (m--) {
    int l, r;
    scanf("%d%d", &l, &r);
    int cnt = 0;
    for (int i = 1; i < top; i++)
      if (dp[i][r] - dp[i][l - 1] == a[i]) cnt++;
    printf("%d\n", cnt);
  }
  return 0;
}
