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;
void init() {}
const int N = (int)1e5 + 34;
int a[N], k[N], o[N];
vector<int> v;
int p[500][N];
int main() {
  init();
  int n, m, l, r;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    if (a[i] < N) k[a[i]]++;
  }
  for (int i = 1; i < N; i++) {
    if (k[i] >= i) {
      v.push_back(i);
    }
  }
  for (int i = 0; i < (int)v.size(); i++) {
    for (int j = 0; j < n; j++) p[i][j + 1] = p[i][j] + int(a[j] == v[i]);
  }
  int ans = 0;
  while (m--) {
    scanf("%d%d", &l, &r);
    ans = 0;
    for (int i = 0; i < (int)v.size(); i++) {
      if (p[i][r] - p[i][l - 1] == v[i]) ans++;
    }
    printf("%d\n", ans);
  }
  return 0;
}
