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;
using VI = vector<int>;
struct Mo {
  int l, r, i;
};
int sqr;
VI A;
VI rep;
int ans = 0;
bool comp(Mo x, Mo y) {
  int bx = x.l / sqr;
  int by = y.l / sqr;
  if (bx != by) return bx < by;
  return (bx & 1 ? x.r > y.r : x.r < y.r);
}
inline void add(int i) {
  int a = A[i];
  if (rep[a] == a) --ans;
  if (++rep[a] == a) ++ans;
}
inline void remove(int i) {
  int a = A[i];
  if (rep[a] == a) --ans;
  if (--rep[a] == a) ++ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, q;
  cin >> n >> q;
  sqr = sqrt(n);
  A = VI(n);
  for (int& a : A) {
    cin >> a;
    if (a > n) a = n + 1;
  }
  vector<Mo> Q(q);
  for (int i = 0; i < q; ++i) {
    int l, r;
    cin >> l >> r;
    Q[i] = {l - 1, r - 1, i};
  }
  sort(Q.begin(), Q.end(), comp);
  rep = VI(n + 2, 0);
  VI res(q);
  int l = 0, r = -1;
  for (Mo m : Q) {
    while (r < m.r) add(++r);
    while (l > m.l) add(--l);
    while (r > m.r) remove(r--);
    while (l < m.l) remove(l++);
    res[m.i] = ans;
  }
  for (int r : res) cout << r << '\n';
}
