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;
int bsize = 300;
struct segm {
  int l, r;
  int lb, rb;
  int id;
};
bool cmp(const segm& s1, const segm& s2) {
  if (s1.lb == s2.lb) return s1.rb < s2.rb;
  return s1.lb < s2.lb;
}
int s = 0;
int n, m;
vector<int> cnt;
void add(int v) {
  if (v > n) return;
  if (cnt[v] == v) s--;
  cnt[v]++;
  if (cnt[v] == v) s++;
}
void del(int v) {
  if (v > n) return;
  if (cnt[v] == v) s--;
  cnt[v]--;
  if (cnt[v] == v) s++;
}
int main() {
  scanf("%d%d", &n, &m);
  cnt.resize(n + 2);
  vector<int> v(n);
  vector<int> ans(m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &v[i]);
  }
  vector<segm> q(m);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    q[i].l = a;
    q[i].r = b;
    q[i].lb = a / bsize;
    q[i].rb = b / bsize;
    q[i].id = i;
  }
  sort(q.begin(), q.end(), cmp);
  int l = 0, r = 0;
  add(v[0]);
  for (int i = 0; i < m; i++) {
    while (l < q[i].l) {
      del(v[l]);
      l++;
    }
    while (l > q[i].l) {
      l--;
      add(v[l]);
    }
    while (r < q[i].r) {
      r++;
      add(v[r]);
    }
    while (r > q[i].r) {
      del(v[r]);
      r--;
    }
    ans[q[i].id] = s;
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
  return 0;
}
