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, a[100001], b[500], dp[100001][500], was[100001];
vector<int> p[100001];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] <= n) p[a[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++)
    if (i <= p[i].size()) {
      b[++b[0]] = i;
      was[i] = b[0];
    }
  if (a[1] <= n && was[a[1]]) dp[1][was[a[1]]] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j < 500; j++) dp[i][j] = dp[i - 1][j];
    if (a[i] <= n && was[a[i]]) dp[i][was[a[i]]]++;
  }
  for (int i = 1; i <= m; i++) {
    int l, r, cnt = 0;
    scanf("%d%d", &l, &r);
    for (int j = 1; j < 500; j++) {
      int k = dp[r][j] - dp[l - 1][j];
      if (was[k] == j) cnt++;
    }
    printf("%d\n", cnt);
  }
  return 0;
}
