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;
int a[100005];
int mp[100005];
int ans[100005], answer;
int sz;
struct data {
  int l, r, ind;
} Q[100005];
bool cmp(data a, data b) {
  if (a.l / sz == b.l / sz) return a.r < b.r;
  return a.l / sz < b.l / sz;
}
void add(int index) {
  mp[a[index]]++;
  if (mp[a[index]] == a[index])
    answer++;
  else if (mp[a[index]] == 1 + a[index])
    answer--;
}
void rem(int index) {
  mp[a[index]]--;
  if (mp[a[index]] == a[index])
    answer++;
  else if (mp[a[index]] + 1 == a[index])
    answer--;
}
int main() {
  int t, n, i, j, cas = 1, q, l, r;
  answer = 0;
  scanf("%d%d", &n, &q);
  sz = sqrt(n);
  memset(mp, 0, sizeof mp);
  for (i = 1; i <= n; i++) {
    scanf("%d", a + i);
    if (a[i] > n) a[i] = -1;
  }
  for (i = 1; i <= q; i++) {
    scanf("%d%d", &Q[i].l, &Q[i].r);
    Q[i].ind = i;
  }
  sort(Q, Q + q, cmp);
  int s = 0, e = 0;
  for (i = 1; i <= q; i++) {
    l = Q[i].l, r = Q[i].r;
    while (s < l) {
      rem(s);
      s++;
    }
    while (s > l) {
      add(s - 1);
      s--;
    }
    while (e <= r) {
      add(e);
      e++;
    }
    while (e > r + 1) {
      rem(e - 1);
      e--;
    }
    ans[Q[i].ind] = answer;
  }
  for (i = 1; i <= q; i++) {
    printf("%d\n", ans[i]);
  }
}
