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;
long long i, n, q, block = 317, L, R = -1, s, a[100001], ans[100001], c[100001];
struct st {
  long long l, r, id;
} qq[100001];
bool cmp(st x, st y) {
  if (x.l / block == y.l / block)
    return x.r < y.r;
  else
    return x.l / block < y.l / block;
}
void add(long long x) {
  if (a[x] <= n) {
    if (++c[a[x]] == a[x])
      s++;
    else if (c[a[x]] == a[x] + 1)
      s--;
  }
}
void rmv(long long x) {
  if (a[x] <= n) {
    if (--c[a[x]] == a[x])
      s++;
    else if (c[a[x]] == a[x] - 1)
      s--;
  }
}
int main() {
  ios_base ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> q;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= q; i++) cin >> qq[i].l >> qq[i].r, qq[i].id = i;
  sort(qq, qq + q, cmp);
  for (i = 1; i <= q; i++) {
    while (L > qq[i].l) add(--L);
    while (L < qq[i].l) rmv(L++);
    while (R > qq[i].r) rmv(R--);
    while (R < qq[i].r) add(++R);
    ans[qq[i].id] = s;
  }
  for (i = 1; i <= q; i++) cout << ans[i] << endl;
}
