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;
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
const int N = 1e5 + 10;
int a[N], f[N], ans[N], cnt[N], l[N], r[N], n, m;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    if (a[i] <= n) cnt[a[i]]++;
  }
  for (int i = 1; i <= m; i++) scanf("%d%d", l + i, r + i);
  for (int i = 1; i <= n; i++) {
    if (cnt[i] >= i) {
      f[0] = 0;
      for (int j = 1; j <= n; j++) f[j] = a[j] != i ? f[j - 1] : f[j - 1] + 1;
      for (int j = 1; j <= m; j++)
        if (f[r[j]] - f[l[j] - 1] == i) ans[j]++;
    }
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}
