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;
struct Query {
  int l, r, id;
} Q[100000];
bool cmp(Query A, Query B) {
  if (A.l / 320 != B.l / 320) {
    return A.l / 320 < B.l / 320;
  }
  return A.r < B.r;
}
int n, m, num;
int a[100000], ans[100000], cnt[100005];
void add(int x) {
  if (a[x] > 100000) return;
  if (cnt[a[x]] == a[x]) {
    num--;
  }
  cnt[a[x]]++;
  if (cnt[a[x]] == a[x]) {
    num++;
  }
}
void del(int x) {
  if (a[x] > 100000) return;
  if (cnt[a[x]] == a[x]) {
    num--;
  }
  cnt[a[x]]--;
  if (cnt[a[x]] == a[x]) {
    num++;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < m; i++) {
    cin >> Q[i].l >> Q[i].r;
    Q[i].l--, Q[i].r--;
    Q[i].id = i;
  }
  sort(Q, Q + m, cmp);
  int L = 0, R = 0;
  for (int i = 0; i < m; i++) {
    while (L < Q[i].l) {
      del(L);
      L++;
    }
    while (L > Q[i].l) {
      add(L - 1);
      L--;
    }
    while (R <= Q[i].r) {
      add(R);
      R++;
    }
    while (R > Q[i].r + 1) {
      del(R - 1);
      R--;
    }
    ans[Q[i].id] = num;
  }
  for (int i = 0; i < m; i++) {
    cout << ans[i] << '\n';
  }
  return 0;
}
