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 MAX = 1e6 + 10;
const int inf = 0x3f3f3f3f;
vector<int> ans[MAX];
int res[MAX], C[MAX], a[MAX];
int n, m;
int lowbit(int x) { return x & -x; }
void modify(int x, int val) {
  for (int i = x; i <= n; i += lowbit(i)) C[i] += val;
}
int sum(int x) {
  int res = 0;
  for (int i = x; i; i -= lowbit(i)) res += C[i];
  return res;
}
struct Query {
  int L, R, id;
  bool operator<(const Query rhs) const { return R < rhs.R; }
} v[MAX];
int main() {
  int x;
  while (scanf("%d %d", &n, &m) == 2) {
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 0; i < m; i++) {
      scanf("%d %d", &v[i].L, &v[i].R);
      v[i].id = i;
    }
    sort(v, v + m);
    int j = 0;
    for (int i = 1; i <= n; i++) {
      if (a[i] <= n) {
        int vv = a[i];
        ans[vv].push_back(i);
        int Size = (int)ans[vv].size();
        if (Size >= vv) {
          modify(ans[vv][Size - vv], 1);
          if (Size > vv) modify(ans[vv][Size - vv - 1], -2);
          if (Size > vv + 1) modify(ans[vv][Size - vv - 2], 1);
        }
      }
      while (j < m && v[j].R == i) {
        res[v[j].id] = sum(v[j].R) - sum(v[j].L - 1);
        j++;
      }
    }
    for (int i = 0; i < m; i++) printf("%d\n", res[i]);
  }
  return 0;
}
