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, ind;
  Query() {}
  Query(int _l, int _r, int _ind) {
    l = _l;
    r = _r;
    ind = _ind;
  }
};
int n, sz;
Query q[100000];
int a[100000];
int ans[100000];
int C = 315;
bool cmp(Query& a, Query& b) {
  int t1 = a.l / C;
  int t2 = b.l / C;
  if (t1 != t2) return t1 < t2;
  return a.r < b.r;
}
int curr = 0;
int mp[100005];
int l = 0;
int r = -1;
void proc(int x, int val) {
  if (x > 100000) return;
  int old = mp[x];
  if (x == old) {
    curr--;
  }
  old += val;
  if (old == x) {
    curr++;
  }
  mp[x] = old;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> sz;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < sz; i++) {
    cin >> q[i].l >> q[i].r;
    q[i].l--;
    q[i].r--;
    q[i].ind = i;
  }
  sort(q, q + sz, cmp);
  for (int i = 0; i < 100005; i++) {
    mp[i] = 0;
  }
  for (int i = 0; i < sz; i++) {
    int nl = q[i].l;
    int nr = q[i].r;
    while (r < nr) {
      int x = a[++r];
      proc(x, 1);
    }
    while (l > nl) {
      int x = a[--l];
      proc(x, 1);
    }
    while (r > nr) {
      int x = a[r--];
      proc(x, -1);
    }
    while (l < nl) {
      int x = a[l++];
      proc(x, -1);
    }
    ans[q[i].ind] = curr;
  }
  for (int i = 0; i < sz; i++) {
    cout << ans[i] << "\n";
  }
  return 0;
}
