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 MOD = 1e9 + 7;
const int MAX = 1e5;
const int MV = 1e5;
const long double PI = 3.14159265358979323846;
int PO(int x, int y) {
  if (y == 0) return 1;
  int ans = PO(x, y / 2);
  ans *= ans, ans %= MOD;
  if (y % 2 == 1) ans *= x, ans %= MOD;
  return ans % MOD;
}
int amo[MAX + 3], dp[MAX + 3][500], arr[MAX + 3];
int n, m;
vector<int> vec;
int32_t main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0),
      cout << fixed << setprecision(10);
  cin >> n >> m;
  int x;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    x = arr[i];
    if (x <= n) {
      amo[x]++;
      if (amo[x] == x) vec.push_back(x);
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < vec.size(); j++) {
      dp[i][j] = dp[i - 1][j];
      if (arr[i] == vec[j]) dp[i][j]++;
    }
  }
  int l, r, ans;
  while (m--) {
    l, r;
    cin >> l >> r;
    ans = 0;
    for (int j = 0; j < vec.size(); j++) {
      if (dp[r][j] - dp[l - 1][j] == vec[j]) ans++;
    }
    cout << ans << "\n";
  }
  return 0;
}
