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 N = (int)1e5 + 123;
int n, Q, block;
int cnt[N], a[N], ans[N];
struct st {
  int l, r, id;
};
vector<st> q;
bool cmp(st f, st s) {
  if (f.l / block != s.l / block) return f.l < s.l;
  return f.r < s.r;
}
int kol = 0;
void add(int pos) {
  pos = a[pos];
  if (pos <= n) {
    cnt[pos]++;
    if (cnt[pos] == pos) kol++;
    if (cnt[pos] == pos + 1) kol--;
  }
}
void del(int pos) {
  pos = a[pos];
  if (pos <= n) {
    cnt[pos]--;
    if (cnt[pos] == pos) kol++;
    if (cnt[pos] == pos - 1) kol--;
  }
}
int main() {
  cin >> n >> Q;
  block = sqrt(n);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= Q; i++) {
    int l, r;
    cin >> l >> r;
    q.push_back({l, r, i});
  }
  sort(q.begin(), q.end(), &cmp);
  int l = 1, r = 0;
  for (int i = 0; i < Q; i++) {
    while (r < q[i].r) add(++r);
    while (l < q[i].l) del(l++);
    while (r > q[i].r) del(r--);
    while (l > q[i].l) add(--l);
    ans[q[i].id] = kol;
  }
  for (int i = 1; i <= Q; i++) cout << ans[i] << endl;
  return 0;
}
