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 ans, n, q;
long long cnt[100005], a[100005], anss[100005], s[100005];
struct node {
  long long l, r, id, block;
} p[100005];
bool cmp(node a, node b) {
  if (a.block == b.block) return a.r < b.r;
  return a.block < b.block;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
void add(long long pos) {
  cnt[a[pos]]++;
  if (cnt[a[pos]] == s[a[pos]]) ans++;
  if (cnt[a[pos]] == s[a[pos]] + 1) ans--;
}
void del(long long pos) {
  cnt[a[pos]]--;
  if (cnt[a[pos]] == s[a[pos]]) ans++;
  if (cnt[a[pos]] == s[a[pos]] - 1) ans--;
}
int main() {
  while (~scanf("%lld%lld", &n, &q)) {
    long long pp = sqrt(n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &s[i]);
      a[i] = s[i];
    }
    sort(s + 1, s + n + 1);
    unique(s + 1, s + n + 1);
    for (long long i = 1; i <= q; i++) {
      long long p = lower_bound(s + 1, s + n + 1, a[i]) - s;
      a[i] = p;
    }
    for (long long i = 1; i <= q; i++) {
      scanf("%lld%lld", &p[i].l, &p[i].r);
      p[i].id = i;
      p[i].block = p[i].l / pp;
    }
    long long L = 1, R = 0;
    sort(p + 1, p + 1 + q, cmp);
    ans = 0;
    for (long long i = 1; i <= q; i++) {
      while (p[i].r > R) add(++R);
      while (p[i].r < R) del(R--);
      while (p[i].l < L) add(--L);
      while (p[i].l > L) del(L++);
      anss[p[i].id] = ans;
    }
    for (long long i = 1; i <= q; i++) {
      printf("%lld\n", anss[i]);
    }
  }
  return 0;
}
