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 BLOCK;
struct query {
  int l, r, id;
};
bool operator<(query a, query b) {
  if (a.l / BLOCK != b.l / BLOCK) return a.l / BLOCK < b.l / BLOCK;
  return a.r < b.r;
}
int vis[100100], resp, vet[100100];
unordered_map<int, int> mp;
void add(int id) {
  if (!vis[id]) {
    if (mp[vet[id]] == vet[id]) resp--;
    mp[vet[id]]++;
    if (mp[vet[id]] == vet[id]) resp++;
  } else {
    if (mp[vet[id]] == vet[id]) resp--;
    mp[vet[id]]--;
    if (mp[vet[id]] == vet[id]) resp++;
  }
  vis[id] ^= 1;
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  BLOCK = 320;
  query queries[q];
  int ans[q];
  for (int i = 0; i < n; i++) scanf("%d", &vet[i]);
  for (int i = 0; i < q; i++)
    scanf("%d %d", &queries[i].l, &queries[i].r), queries[i].id = i;
  sort(queries, queries + q);
  int curL = 0, curR = 0;
  resp = 0;
  for (int i = 0; i < q; i++) {
    int l = queries[i].l - 1, r = queries[i].r - 1;
    while (curL < l) add(curL++);
    while (curL > l) add(--curL);
    while (curR <= r) add(curR++);
    while (curR > r + 1) add(--curR);
    ans[queries[i].id] = resp;
  }
  for (int i = 0; i < q; i++) printf("%d\n", ans[i]);
  return 0;
}
