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 ri() {
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  int ret = 0;
  while (c >= '0' && c <= '9') {
    ret = 10 * ret + c - 48;
    c = getchar();
  }
  return ret;
}
int n, m, a[100010], c[501], dp[501][100010], cnt, u, v;
map<int, int> mo;
int main() {
  cin >> n >> m;
  for (long long i = (1); (i) < (n + 1); ++i) {
    cin >> a[i];
    mo[a[i]]++;
    if (mo[a[i]] == a[i]) c[++cnt] = a[i];
  }
  for (long long i = (1); (i) < (cnt + 1); ++i) {
    for (long long j = (1); (j) < (n + 1); ++j) {
      dp[i][j] = dp[i][j - 1];
      if (a[j] == c[i]) dp[i][j]++;
    }
  }
  for (long long i = (1); (i) < (m + 1); ++i) {
    cin >> u >> v;
    int ans = 0;
    for (long long j = (1); (j) < (cnt + 1); ++j) {
      if (dp[j][v] - dp[j][u - 1] == c[j]) ans++;
    }
    cout << ans << "\n";
  }
}
