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 n, k, Q, S, res, a[100005], cnt[100005], f[100005], block[100005];
struct query {
  long long l, r, index;
} q[100005];
bool cmp(query A, query B) {
  if (block[A.l] != block[B.l]) return block[A.l] < block[B.l];
  return A.r < B.r;
}
void add_(long long pos) {
  if (a[pos] > n) return;
  cnt[a[pos]]++;
  if (cnt[a[pos]] == a[pos] + 1) res--;
  if (cnt[a[pos]] == a[pos]) res++;
}
void remove_(long long pos) {
  if (a[pos] > n) return;
  cnt[a[pos]]--;
  if (cnt[a[pos]] == a[pos] - 1) res--;
  if (cnt[a[pos]] == a[pos]) res++;
}
int main() {
  scanf("%lld%lld", &n, &Q);
  S = sqrt(n);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    block[i] = i / S;
  }
  for (long long i = 1; i <= Q; i++) {
    scanf("%lld%lld", &q[i].l, &q[i].r);
    q[i].index = i;
  }
  sort(q + 1, q + 1 + Q, cmp);
  long long curL = 0, curR = 0;
  for (long long i = 1; i <= Q; i++) {
    while (curL < q[i].l) {
      remove_(curL);
      curL++;
    }
    while (curL > q[i].l) {
      curL--;
      add_(curL);
    }
    while (curR < q[i].r) {
      curR++;
      add_(curR);
    }
    while (curR > q[i].r) {
      remove_(curR);
      curR--;
    }
    f[q[i].index] = res + 1;
  }
  for (long long i = 1; i <= Q; i++) printf("%lld\n", f[i]);
}
