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 INF = 2147483647;
const int MOD = (int)1e9 + 7;
const int MAX = (int)1e5 + 5;
const int MAXN = 460;
int stat[MAXN][MAX], a[MAX], cnt[MAX];
vector<int> cand;
int main() {
  int n, m;
  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 <= n; i++) {
    if (cnt[i] >= i) cand.push_back(i);
  }
  for (int(i) = 0; (i) < (int)(((int)(cand).size())); (i)++) {
    for (int j = 1; j <= n; j++) {
      if (cand[i] == a[j])
        stat[i][j] = stat[i][j - 1] + 1;
      else
        stat[i][j] = stat[i][j - 1];
    }
  }
  while (m--) {
    int l, r;
    scanf("%d%d", &l, &r);
    int ans = 0;
    for (int(i) = 0; (i) < (int)(((int)(cand).size())); (i)++) {
      if (stat[i][r] - stat[i][l - 1] == cand[i]) ans++;
    }
    printf("%d\n", ans);
  }
  return 0;
}
