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 = 100010;
map<int, int> M;
vector<int> adj[MAXN];
vector<int> tadj[MAXN];
int n, m, l, r, vn, tvn, a[MAXN], v[MAXN], tv[MAXN];
int main() {
  while (scanf("%d%d", &n, &m) != EOF) {
    M.clear();
    for (int i = 0; i < n; ++i) adj[i].clear();
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
      M[a[i]] = 0;
    }
    vn = 0;
    for (map<int, int>::iterator it = M.begin(); it != M.end(); ++it) {
      v[vn] = it->first;
      it->second = vn++;
    }
    for (int i = 1; i <= n; ++i) {
      adj[M[a[i]]].push_back(i);
    }
    tvn = 0;
    for (int i = 0; i < vn; ++i) {
      if (adj[i].size() >= v[i]) {
        tv[tvn] = v[i];
        tadj[tvn].clear();
        tadj[tvn].insert(tadj[tvn].end(), adj[i].begin(), adj[i].end());
        ++tvn;
      }
    }
    while (m--) {
      scanf("%d%d", &l, &r);
      int ans = 0;
      for (int i = 0; i < tvn; ++i) {
        vector<int>::iterator it =
            lower_bound(tadj[i].begin(), tadj[i].end(), l);
        if (it != tadj[i].end()) {
          int ii = it - tadj[i].begin();
          if (ii + tv[i] - 1 < tadj[i].size() &&
              (tadj[i][ii + tv[i] - 1] <= r) &&
              (ii + tv[i] >= tadj[i].size() || tadj[i][ii + tv[i]] > r)) {
            ++ans;
          }
        }
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
