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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
int power(int x, unsigned int y) {
  int res = 1;
  while (y > 0) {
    if (y & 1) {
      res = res * x;
    }
    y = y >> 1;
    x = x * x;
  }
  return res;
}
int powermod(int x, unsigned int y, int p) {
  int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % p;
    }
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int Mod(int x, int m) { return (x % m + m) % m; }
int min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }
const int mxn = 1e5 + 5;
int freq[mxn];
int arr[mxn];
int n;
int bs = 300;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] > 100000) {
      arr[i] = 0;
    }
  }
  vector<pair<pair<int, int>, int>> v;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    v.push_back(make_pair(make_pair(a, b), i));
  }
  sort(v.begin(), v.end(),
       [](pair<pair<int, int>, int> &a, pair<pair<int, int>, int> &b) -> bool {
         if (a.first.first / bs != b.first.first / bs)
           return a.first.first < b.first.first;
         else {
           return a.first.second < b.first.second;
         }
       });
  int l = 0;
  int r = -1;
  for (int i = 0; i < mxn; i++) {
    freq[i] = 0;
  }
  int ans = 0;
  freq[0] = 1000;
  vector<int> res(m + 1);
  for (int i = 0; i < m; i++) {
    int low = v[i].first.first;
    int high = v[i].first.second;
    while (l < low) {
      if (freq[arr[l]] == arr[l]) {
        ans--;
      }
      freq[arr[l]]--;
      if (freq[arr[l]] == arr[l]) {
        ans++;
      }
      l++;
    }
    while (l > low) {
      l--;
      if (freq[arr[l]] == arr[l]) {
        ans--;
      }
      freq[arr[l]]++;
      if (freq[arr[l]] == arr[l]) {
        ans++;
      }
    }
    while (r < high) {
      r++;
      if (freq[arr[r]] == arr[r]) {
        ans--;
      }
      freq[arr[r]]++;
      if (freq[arr[r]] == arr[r]) {
        ans++;
      }
    }
    while (r > high) {
      if (freq[arr[r]] == arr[r]) {
        ans--;
      }
      freq[arr[r]]--;
      if (freq[arr[r]] == arr[r]) {
        ans++;
      }
      r--;
    }
    res[v[i].second] = ans;
  }
  for (int i = 0; i < m; i++) {
    cout << res[i] << '\n';
  }
  return 0;
}
