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 + 100;
int n, m;
int a[MAXN], c[MAXN], ans[MAXN];
struct Query {
  int l, r, id;
  bool operator<(const Query &t) const { return r < t.r; }
} q[MAXN];
inline int lowbit(int x) { return x & (-x); }
void add(int i, int v) {
  while (i <= n) {
    c[i] += v;
    i += lowbit(i);
  }
}
int sum(int x) {
  int ret = 0;
  while (x > 0) {
    ret += c[x];
    x -= lowbit(x);
  }
  return ret;
}
int main() {
  int sz;
  while (~scanf("%d%d", &n, &m)) {
    vector<int> data[MAXN];
    memset(c, 0, sizeof(c));
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= m; i++) {
      scanf("%d%d", &q[i].l, &q[i].r);
      q[i].id = i;
    }
    sort(q + 1, q + 1 + m);
    for (int i = 1, k = 1; i <= n; i++) {
      if (a[i] <= n) {
        data[a[i]].push_back(i);
        sz = data[a[i]].size();
        if (sz >= a[i]) {
          add(data[a[i]][sz - a[i]], 1);
          if (sz > a[i]) add(data[a[i]][sz - a[i] - 1], -2);
          if (sz > a[i] + 1) add(data[a[i]][sz - a[i] - 2], 1);
        }
      }
      while (q[k].r == i && k <= m) {
        ans[q[k].id] = sum(q[k].r) - sum(q[k].l - 1);
        k++;
      }
    }
    for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  }
  return 0;
}
