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 = 1e9;
const int MAXN = 100002;
const int MAXR = 400;
int a[MAXN];
int c[MAXN];
int cb = 0;
int b[MAXN];
int cand[2 * MAXR + 1];
int pr[600][MAXN + 1];
int cc = 0;
int main() {
  int n, m;
  cin >> n >> m;
  memset(c, 0, sizeof(c));
  for (int i = 0; i < (n); ++i) {
    scanf("%d", &a[i]);
    if (a[i] <= n) c[a[i]]++;
  }
  for (int i = 1; i < n + 1; ++i) {
    if (c[i] >= i) cand[cc++] = i;
  }
  memset(c, 0, sizeof(c));
  memset(pr, 0, sizeof(pr));
  for (int i = 0; i < (n); ++i) {
    for (int j = 0; j < (cc); ++j)
      if (a[i] == cand[j]) c[j]++;
    for (int j = 0; j < (cc); ++j) {
      pr[j][i + 1] = c[j];
    }
  }
  for (int i = 0; i < (m); ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    int ans = 0;
    for (int j = 0; j < (cc); ++j) {
      int w = pr[j][r] - pr[j][l - 1];
      if (w == cand[j]) ans++;
    }
    printf("%d\n", ans);
  }
  return 0;
}
