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 N = 1e5 + 5;
const int BLOCK = 320;
struct Query {
  int l, r, id;
};
Query q[N];
int ans, a[N], prin[N];
unordered_map<int, int> cnt;
bool cmp(Query x, Query y) {
  return x.l / BLOCK == y.l / BLOCK ? x.r < y.r : x.l / BLOCK < y.l / BLOCK;
}
void add(int p) {
  if (cnt[a[p]] == a[p] && p > 0) ans--;
  cnt[a[p]]++;
  if (cnt[a[p]] == a[p] && p > 0) ans++;
}
void del(int p) {
  if (cnt[a[p]] == a[p] && p > 0) ans--;
  cnt[a[p]]--;
  if (cnt[a[p]] == a[p] && p > 0) ans++;
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 1; 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].id = i;
  }
  sort(q, q + m, cmp);
  int l = 1, r = 0;
  for (int i = 0; i < m; i++) {
    while (r < q[i].r) add(++r);
    while (l > q[i].l) add(--l);
    while (l < q[i].l) del(l++);
    while (r > q[i].r) del(r--);
    prin[q[i].id] = ans;
  }
  for (int i = 0; i < m; i++) {
    printf("%d\n", prin[i]);
  }
  return 0;
}
