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[100002], b[100002][320], ls[100002], o;
int c[100002][320];
map<int, int> M, M2;
int main() {
  cin >> n >> m;
  int t = sqrt(n) + 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    M[a[i]]++;
    for (int j = 1; j <= t; j++) b[i][j] = b[i - 1][j] + (a[i] == j);
  }
  for (int i = 1; i <= n; i++)
    if (a[i] > t && M[a[i]] > t && !M2[a[i]]) {
      ls[++o] = a[i];
      M2[a[i]] = 1;
    }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= o; j++) c[i][j] = c[i - 1][j] + (a[i] == ls[j]);
  }
  for (int l, r; m--;) {
    scanf("%d%d", &l, &r);
    long long as = 0;
    for (int i = 1; i <= t; i++) as += b[r][i] - b[l - 1][i] == i;
    for (int i = 1; i <= o; i++) as += c[r][i] - c[l - 1][i] == ls[i];
    printf("%lld\n", as);
  }
  return 0;
}
