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;
const int maxn = 100007;
int sz = 316;
struct Node {
  int l, r, id;
  friend bool operator<(Node a, Node b) {
    return (a.l / sz) == (b.l / sz) ? a.r < b.r : a.l < b.l;
  }
} q[maxn];
int pos[maxn];
int sum[maxn];
int a[maxn];
int ans[maxn];
unordered_map<int, int> mp;
int Ans = 0;
int n, m, k;
void add(int pos) {
  if (mp[a[pos]] == a[pos]) {
    mp[a[pos]]++;
    Ans--;
  } else {
    mp[a[pos]]++;
    if (mp[a[pos]] == a[pos]) Ans++;
  }
}
void del(int pos) {
  if (mp[a[pos]] == a[pos]) {
    mp[a[pos]]--;
    Ans--;
  } else {
    mp[a[pos]]--;
    if (mp[a[pos]] == a[pos]) Ans++;
  }
}
int L = 1, R = 0;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &q[i].l, &q[i].r);
    q[i].id = i;
  }
  sort(q + 1, q + 1 + m);
  for (int i = 1; i <= m; i++) {
    while (L < q[i].l) {
      del(L);
      L++;
    }
    while (L > q[i].l) {
      L--;
      add(L);
    }
    while (R < q[i].r) {
      R++;
      add(R);
    }
    while (R > q[i].r) {
      del(R);
      R--;
    }
    ans[q[i].id] = Ans;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
}
