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;
int bs;
int* cnt;
int ans;
class query {
 public:
  int l;
  int r;
  int index;
};
bool cmp(const query a, const query b) {
  int s = a.l / bs;
  int e = b.l / bs;
  if (s == e) return a.r < b.r;
  return s < e;
}
void add(int x) {
  if (cnt[x] == x)
    ans--;
  else if (cnt[x] == x - 1)
    ans++;
  cnt[x]++;
}
void del(int x) {
  if (cnt[x] == x)
    ans--;
  else if (cnt[x] == x + 1)
    ans++;
  cnt[x]--;
}
int main() {
  int n, m, *a;
  query* q;
  scanf("%d %d", &n, &m);
  bs = sqrt(n);
  a = (int*)calloc(n + 1, sizeof(int));
  cnt = (int*)calloc(n + 1, sizeof(int));
  int* an = (int*)calloc(m + 1, sizeof(int));
  q = (query*)calloc(m + 1, sizeof(query));
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &q[i].l, &q[i].r);
    q[i].index = i;
    q[i].l--;
    q[i].r--;
  }
  sort(q, q + m, cmp);
  ans = 0;
  int l = 0, r = -1;
  for (int i = 0; i < m; i++) {
    int left = q[i].l;
    int right = q[i].r;
    while (r < right) {
      r++;
      if (a[r] < n + 1) add(a[r]);
    }
    while (r > right) {
      if (a[r] < n + 1) del(a[r]);
      r--;
    }
    while (l < left) {
      if (a[l] < n + 1) del(a[l]);
      l++;
    }
    while (l > left) {
      l--;
      if (a[l] < n + 1) add(a[l]);
    }
    an[q[i].index] = ans;
  }
  for (int i = 0; i < m; i++) printf("%d\n", an[i]);
  free(a);
  free(an);
  free(cnt);
  free(q);
  return 0;
}
