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 = 100005;
int blocksize;
int ar[N], res[N], ans, L, R, n, cnt[N];
struct Query {
  int i, j, id;
  bool operator<(const Query& q) const {
    if (i / blocksize != q.i / blocksize) {
      return i < q.i;
    }
    return j > q.j;
  }
};
Query qr[N];
inline void add(int i) {
  if (ar[i] > n) return;
  if (ar[i] == cnt[ar[i]]) ans--;
  cnt[ar[i]]++;
  ans += (ar[i] == (cnt[ar[i]]));
}
inline void del(int i) {
  if (ar[i] > n) return;
  if (ar[i] == cnt[ar[i]]) ans--;
  cnt[ar[i]]--;
  ans += (ar[i] == (cnt[ar[i]]));
}
int main() {
  int m;
  scanf("%d %d", &n, &m);
  blocksize = sqrt(n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &ar[i]);
  }
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &qr[i].i, &qr[i].j);
    qr[i].id = i;
  }
  sort(qr, qr + m);
  L = R = 1;
  if (ar[1] <= n) cnt[ar[1]]++;
  ans += (cnt[1] == 1);
  for (int i = 0; i < m; i++) {
    int l = qr[i].i, r = qr[i].j;
    while (l < L) add(--L);
    while (l > L) del(L++);
    while (r < R) del(R--);
    while (r > R) add(++R);
    res[qr[i].id] = ans;
  }
  for (int i = 0; i < m; i++) {
    printf("%d\n", res[i]);
  }
  return 0;
}
