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 long long maxn = 1e5 + 10, maxt = 1e9 + 10, inf = 1e15, radical = 500,
                mod = 29959;
inline int sc() {
  int x;
  scanf("%d", &x);
  return x;
}
vector<int> v[maxn];
inline int bs(int x, int id) {
  int fi = 0, se = v[id].size() - 1;
  while (se - fi > 1) {
    int mid = (fi + se) / 2;
    if (v[id][mid] <= x)
      fi = mid;
    else
      se = mid;
  }
  return fi;
}
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) v[i].push_back(0);
  for (int i = 1; i <= n; i++) {
    int x;
    x = sc();
    if (x <= n) v[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) v[i].push_back(n + 1);
  vector<int> k;
  for (int i = 1; i <= n; i++)
    if (v[i].size() - 2 >= i) k.push_back(i);
  while (m--) {
    int l, r;
    l = sc();
    r = sc();
    int ans = 0;
    for (int i = 0; i < k.size(); i++) {
      int x = k[i];
      if (bs(r, x) - (lower_bound(v[x].begin(), v[x].end(), l) - v[x].begin()) +
              1 ==
          x)
        ans++;
    }
    printf("%d\n", ans);
  }
  return 0;
}
