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 = 1e5 + 5, M = 1e5 + 5;
int n, m, qs[M], qe[M], sorted[M], arr[N], ans, s, e, answers[M];
unordered_map<int, int> freq;
void add(int val) {
  if (freq[val] == val) ans--;
  freq[val]++;
  if (freq[val] == val) ans++;
}
void remove(int val) {
  if (freq[val] == val) ans--;
  --freq[val];
  if (freq[val] == val) ans++;
}
void update(int idx) {
  while (s > qs[idx]) add(arr[--s]);
  while (e < qe[idx]) add(arr[e++]);
  while (s < qs[idx]) remove(arr[s++]);
  while (e > qe[idx]) remove(arr[--e]);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", arr + i);
  }
  int l, r;
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &l, &r);
    qs[i] = --l, qe[i] = r;
    sorted[i] = i;
  }
  int sq = sqrt(n);
  sort(sorted, sorted + m, [sq](int a, int b) {
    return make_pair(qs[a] / sq, qe[a]) < make_pair(qs[b] / sq, qe[b]);
  });
  for (int i = 0; i < m; i++) {
    update(sorted[i]);
    answers[sorted[i]] = ans;
  }
  for (int i = 0; i < m; i++) printf("%d\n", answers[i]);
  return 0;
}
