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, num[100005];
int top, to[100005];
int sum[100005][500];
map<int, int> mp;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &num[i]);
    mp[num[i]]++;
  }
  for (map<int, int>::iterator it = mp.begin(); it != mp.end(); it++)
    if (it->first <= it->second) to[top++] = it->first;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j < top; ++j) sum[i][j] = sum[i - 1][j] + (num[i] == to[j]);
  for (int i = 0; i < m; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    int ans = 0;
    for (int j = 0; j < top; j++) ans += (sum[r][j] - sum[l - 1][j]) == to[j];
    printf("%d\n", ans);
  }
}
