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 N = 100005;
int sum[N];
void update(int pos, int delta) {
  for (; pos < N; pos += pos & -pos) sum[pos] += delta;
}
int query(int pos) {
  int ans = 0;
  for (; pos > 0; pos -= pos & -pos) ans += sum[pos];
  return ans;
}
int query(int le, int ri) { return query(ri) - query(le - 1); }
int a[N], pos[N];
vector<int> adj[N], queries[N], id[N];
int ans[N];
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    if (a[i] < N) adj[a[i]].push_back(i);
  }
  int x, y;
  for (int i = 0; i < m; ++i) {
    scanf("%d %d", &x, &y);
    queries[y].push_back(x);
    id[y].push_back(i);
  }
  for (int i = 1; i <= n; ++i) {
    if (a[i] < N) {
      int sz = pos[a[i]] + 1;
      if (sz >= a[i]) {
        update(adj[a[i]][sz - a[i]], +1);
        if (sz >= a[i] + 1) update(adj[a[i]][sz - a[i] - 1], -2);
        if (sz > a[i] + 1) update(adj[a[i]][sz - a[i] - 2], +1);
      }
      pos[a[i]]++;
    }
    for (int j = 0; j < queries[i].size(); ++j) {
      ans[id[i][j]] = query(queries[i][j], i);
    }
  }
  for (int i = 0; i < m; ++i) printf("%d\n", ans[i]);
  return 0;
}
