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 M = 100005;
int n, m, sz, a[M], cnt[M], t, ans[M];
struct node {
  int le, ri, b, id;
} p[M];
bool cmp(node u, node v) {
  if (u.b != v.b) return u.b < v.b;
  return u.ri < v.ri;
}
void add(int x) {
  if (a[x] > n) return;
  if (cnt[a[x]] == a[x]) t--;
  cnt[a[x]]++;
  if (cnt[a[x]] == a[x]) t++;
}
void del(int x) {
  if (a[x] > n) return;
  if (cnt[a[x]] == a[x]) t--;
  cnt[a[x]]--;
  if (cnt[a[x]] == a[x]) t++;
}
int main() {
  scanf("%d%d", &n, &m);
  sz = int(sqrt(n));
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &p[i].le, &p[i].ri);
    p[i].b = p[i].le / sz;
    p[i].id = i;
  }
  sort(p + 1, p + 1 + m, cmp);
  int l = 0, r = 0;
  for (int i = 1; i <= m; i++) {
    while (r < p[i].ri) add(++r);
    while (r > p[i].ri) del(r--);
    while (l > p[i].le - 1) add(l--);
    while (l < p[i].le - 1) del(++l);
    ans[p[i].id] = t;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}
