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, MAXC = 500;
int can[MAXC], c = 0;
int n, m, a[MAXN], b[MAXN];
int t[MAXN][MAXC];
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  memcpy(b, a, sizeof(b));
  sort(b, b + n);
  for (int i = 0; i < n; i++) {
    int end = i;
    while (b[end] == b[i]) end++;
    if (end - i >= b[i]) can[c++] = b[i];
    i = end - 1;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < c; j++) {
      t[i + 1][j] = t[i][j];
      if (a[i] == can[j]) t[i + 1][j]++;
    }
  }
  int l, r;
  for (int i = 0; i < m; i++) {
    cin >> l >> r;
    int ans = 0;
    for (int j = 0; j < c; j++)
      if (t[r][j] - t[l - 1][j] == can[j]) ans++;
    cout << ans << endl;
  }
  return 0;
}
