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 long long maxn = 1e5 + 1, k = 500;
int a[maxn], n, q, dd[maxn], cnt[maxn][k];
vector<int> b;
void read() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
}
void process() {
  for (int i = 1; i <= n; i++)
    if (a[i] <= n) dd[a[i]]++;
  for (int i = 1; i <= n; i++)
    if (dd[i] >= i) b.push_back(i);
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < b.size(); j++)
      if (b[j] == a[i])
        cnt[i][j] = cnt[i - 1][j] + 1;
      else
        cnt[i][j] = cnt[i - 1][j];
  int l, r;
  for (int i = 1; i <= q; i++) {
    cin >> l >> r;
    int counted = 0;
    for (int j = 0; j < b.size(); j++)
      if (cnt[r][j] - cnt[l - 1][j] == b[j]) counted++;
    cout << counted << endl;
  }
}
int main() {
  read();
  process();
  return 0;
}
