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 BS;
int n, m;
int a[100001];
int order[100001];
int c[100001];
int out[100001];
pair<int, int> q[100001];
int ans;
bool cmp(int a, int b) {
  if (q[a].first / BS != q[b].first / BS)
    return q[a].first / BS < q[b].first / BS;
  return q[a].second < q[b].second;
}
void change(int val, int d) {
  if (val <= n) {
    if (c[val] == val) {
      ans -= 1;
    }
    c[val] += d;
    if (c[val] == val) {
      ans += 1;
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  BS = sqrt(n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &q[i].first, &q[i].second);
    order[i] = i;
  }
  sort(order, order + m, cmp);
  int ml = 1;
  int mr = 0;
  for (int i = 0; i < m; i++) {
    int l = q[order[i]].first;
    int r = q[order[i]].second;
    while (mr < r) {
      mr++;
      change(a[mr], 1);
    }
    while (mr > r) {
      change(a[mr], -1);
      mr--;
    }
    while (ml < l) {
      change(a[ml], -1);
      ml++;
    }
    while (ml > l) {
      ml--;
      change(a[ml], 1);
    }
    out[order[i]] = ans;
  }
  for (int i = 0; i < m; i++) printf("%d\n", out[i]);
  return 0;
}
