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 idx, l, r;
} qr[100005];
int sum, sq, ar[100005], ans[100005];
unordered_map<int, int> mp;
bool cmp(Query A, Query B) {
  int a, b;
  a = A.l / sq, b = B.l / sq;
  if (a == b) return A.r < B.r;
  return a < b;
}
void Add(int x) {
  if (ar[x] == mp[ar[x]]) sum--;
  mp[ar[x]]++;
  if (ar[x] == mp[ar[x]]) sum++;
  return;
}
void Remove(int x) {
  if (ar[x] == mp[ar[x]]) sum--;
  mp[ar[x]]--;
  if (ar[x] == mp[ar[x]]) sum++;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, q, l, r;
  cin >> n >> q;
  sq = sqrt(n);
  for (int i = 1; i <= n; i++) cin >> ar[i];
  for (int i = 1; i <= q; i++) {
    cin >> l >> r;
    qr[i].idx = i;
    qr[i].l = l;
    qr[i].r = r;
  }
  sort(qr + 1, qr + 1 + q, cmp);
  l = 1, r = 0;
  for (int i = 1; i <= q; i++) {
    while (r < qr[i].r) Add(++r);
    while (l < qr[i].l) Remove(l++);
    while (r > qr[i].r) Remove(r--);
    while (l > qr[i].l) Add(--l);
    ans[qr[i].idx] = sum;
  }
  for (int i = 1; i <= q; i++) cout << ans[i] << "\n";
  return 0;
}
