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 int oo = (int)1e9;
const double PI = 2 * acos(0.0);
const double eps = 1e-9;
int dcmp(double a, double b) {
  return (fabs(a - b) <= eps) ? 0 : ((a < b) ? -1 : 1);
}
int arr[100100];
int cnt[100100];
int ans[100100];
int cum[100100];
bool done[100100];
pair<int, int> qu[100100];
int n, q;
int main() {
  cin >> n >> q;
  vector<int> good;
  memset(cnt, 0, sizeof cnt);
  memset(done, 0, sizeof done);
  memset(cum, 0, sizeof cum);
  for (int i = 1; i <= n; ++i) {
    cin >> arr[i];
    if (arr[i] <= n) {
      cnt[arr[i]]++;
      if (cnt[arr[i]] >= arr[i] and !done[arr[i]]) {
        done[arr[i]] = 1;
        good.push_back(arr[i]);
      }
    }
  }
  for (int i = 0; i < q; ++i) cin >> qu[i].first >> qu[i].second;
  for (int i = 0; i < (int)good.size(); ++i) {
    int cur = good[i];
    for (int i = 1; i <= n; ++i) cum[i] = cum[i - 1] + (arr[i] == cur);
    for (int k = 0; k < q; ++k) {
      if (cum[qu[k].second] - cum[qu[k].first - 1] == cur) ans[k]++;
    }
  }
  for (int i = 0; i < q; ++i) {
    cout << ans[i] << endl;
  }
  return 0;
}
