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 a[1000000];
int main() {
  vector<int> vec;
  int n, q;
  cin >> n >> q;
  int i, j;
  for (i = 0; i < n; i++) cin >> a[i];
  unordered_map<int, int> m;
  for (i = 0; i < n; i++) {
    if (m.count(a[i]) > 0)
      m[a[i]]++;
    else
      m[a[i]] = 1;
    if (m[a[i]] == a[i]) vec.push_back(a[i]);
  }
  sort(vec.begin(), vec.end());
  int nn = vec.size();
  int c[n][nn];
  int p, l, r, mid, ans;
  for (i = 0; i < n; i++) {
    for (j = 0; j < nn; j++) c[i][j] = 0;
  }
  for (i = 0; i < n; i++) {
    l = 0;
    r = nn - 1;
    ans = -1;
    while (l <= r) {
      mid = (l + r) / 2;
      if (vec[mid] == a[i]) {
        ans = mid;
        break;
      }
      if (vec[mid] > a[i])
        r = mid - 1;
      else
        l = mid + 1;
    }
    if (ans != -1) {
      c[i][ans]++;
    }
    if (i > 0) {
      for (j = 0; j < nn; j++) {
        c[i][j] = c[i - 1][j] + c[i][j];
      }
    }
  }
  int f, s, t;
  for (i = 0; i < q; i++) {
    cin >> f >> s;
    t = 0;
    for (j = 0; j < nn; j++) {
      if (f == 1) {
        if (c[s - 1][j] == vec[j]) t++;
      } else {
        if (c[s - 1][j] - c[f - 2][j] == vec[j]) t++;
      }
    }
    cout << t << "\n";
  }
  return 0;
}
