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 MAX_N = 100000 + 100;
int n, query, a[MAX_N], b[MAX_N], cnt[MAX_N][450];
vector<int> p;
int main() {
  scanf("%d %d", &n, &query);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  memcpy(b, a, sizeof b);
  sort(b, b + n);
  for (int i = 0; i < n;) {
    int j = i + 1;
    while (j < n && b[j - 1] == b[j]) j++;
    if (j - i >= b[j - 1]) {
      p.push_back(b[j - 1]);
    }
    i = j;
  }
  memset(cnt, 0, sizeof cnt);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < p.size(); j++) {
      if (p[j] == a[i - 1]) cnt[i][j]++;
      cnt[i][j] += cnt[i - 1][j];
    }
  }
  while (query--) {
    int l, r;
    scanf("%d %d", &l, &r);
    int ans = 0;
    for (int i = 0; i < p.size(); i++) {
      if (p[i] == (cnt[r][i] - cnt[l - 1][i])) ans++;
    }
    printf("%d\n", ans);
  }
}
