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 = 1e6 + 10;
int n, m, l = 1, r, tot, block, a[maxn], sum[maxn], ans[maxn];
struct query {
  int l, r, id;
} q[maxn];
bool cmp(query a, query b) {
  if (a.l / block == b.l / block) return a.r < b.r;
  return a.l < b.l;
}
void add(int x) {
  if (a[x] > n) return;
  if (sum[a[x]] == a[x]) tot--;
  sum[a[x]]++;
  if (sum[a[x]] == a[x]) tot++;
}
void del(int x) {
  if (a[x] > n) return;
  if (sum[a[x]] == a[x]) tot--;
  sum[a[x]]--;
  if (sum[a[x]] == a[x]) tot++;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  block = sqrt(n);
  for (int i = 1; i <= m; i++) scanf("%d%d", &q[i].l, &q[i].r), q[i].id = i;
  sort(q + 1, q + m + 1, cmp);
  for (int i = 1; i <= m; i++) {
    while (l > q[i].l) add(--l);
    while (r < q[i].r) add(++r);
    while (l < q[i].l) del(l++);
    while (r > q[i].r) del(r--);
    ans[q[i].id] = tot;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}
