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 n, m, q, arr[100005];
int mp[100005], f[100005], ans[100005], bit[100005];
vector<int> pos[100005], qr[100005], i[100005];
void update(int i, int v) {
  while (i <= n) {
    bit[i] += v;
    i += i & -i;
  }
}
int query(int i) {
  int r = 0;
  while (i > 0) {
    r += bit[i];
    i -= i & -i;
  }
  return r;
}
int main() {
  scanf("%d %d", &n, &q);
  for (int a = 0; a < n; a++) scanf("%d", &arr[a]);
  for (int a = 0; a < q; a++) {
    int l, r;
    scanf("%d %d", &l, &r);
    qr[r - 1].push_back(l - 1);
    i[r - 1].push_back(a);
  }
  for (int a = 0, b = 0; a < n && b < q; a++) {
    int x = arr[a], y;
    int l = pos[x].size();
    if (x > n) goto hell;
    pos[x].push_back(a), l++;
    if (l >= x) {
      y = pos[x][l - x];
      update(y + 1, 1);
    }
    if (l > x) {
      y = pos[x][l - x - 1];
      update(y + 1, -2);
    }
    if (l > x + 1) {
      y = pos[x][l - x - 2];
      update(y + 1, 1);
    }
  hell:
    int len = qr[a].size();
    for (int j = 0; j < len; j++, b++)
      ans[i[a][j]] = query(a + 1) - query(qr[a][j]);
  }
  for (int a = 0; a < q; a++) printf("%d\n", ans[a]);
  return 0;
}
