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 tab[100005], cou[100005], num[550][100005], n, m, l, r, i, j, res, now, len;
vector<int> pos;
int main() {
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; ++i) {
    scanf("%d", &tab[i]);
    if (tab[i] <= n) ++cou[tab[i]];
  }
  for (i = 1; i <= n; ++i)
    if (cou[i] >= i) pos.push_back(i);
  for (i = 0; i < pos.size(); ++i)
    for (j = 1; j <= n; ++j) {
      num[i][j] = num[i][j - 1];
      if (tab[j] == pos[i]) ++num[i][j];
    }
  for (i = 0; i < m; ++i) {
    scanf("%d%d", &l, &r);
    res = 0;
    len = r - l + 1;
    for (j = 0; j < pos.size(); ++j) {
      if (pos[j] > len) break;
      now = num[j][r] - num[j][l - 1];
      len -= now;
      if (now == pos[j]) ++res;
    }
    printf("%d\n", res);
  }
  return 0;
}
