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 MAXN = 1e5 + 10, SQ = 500;
int n, q, l, r, ans, cnt[MAXN + 5], a[MAXN + 50], dp[MAXN + 50][SQ];
vector<int> v;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] >= MAXN) continue;
    cnt[a[i]]++;
    if (cnt[a[i]] == a[i]) v.push_back(a[i]);
  }
  for (int i = 0; i < v.size(); i++) {
    int u = v[i], tmp = 0;
    for (int j = 1; j <= n; j++) tmp += (a[j - 1] == u), dp[j][i] = tmp;
  }
  for (int i = 0; i < q; i++) {
    cin >> l >> r;
    ans = 0;
    for (int j = 0; j < v.size(); j++) ans += (v[j] == dp[r][j] - dp[l - 1][j]);
    cout << ans << '\n';
  }
}
