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, idx;
};
int n, q;
int a[100010];
const int sq = sqrt(100000) + 1;
int ct[100010], ans = 0;
vector<query> seg;
int answer[100010];
bool cmp(query a, query b) {
  if (a.l / sq == b.l / sq) return a.r < b.r;
  return (a.l / sq < b.l / sq);
}
void add(int x) {
  if (a[x] > 1e5) return;
  if (ct[a[x]] == a[x]) ans--;
  ct[a[x]]++;
  if (ct[a[x]] == a[x]) ans++;
}
void del(int x) {
  if (a[x] > 1e5) return;
  if (ct[a[x]] == a[x]) ans--;
  ct[a[x]]--;
  if (ct[a[x]] == a[x]) ans++;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  while (q--) {
    int a, b;
    scanf("%d%d", &a, &b);
    int sz = seg.size();
    seg.emplace_back(query({a, b, sz}));
  }
  sort(seg.begin(), seg.end(), cmp);
  int ls, rs;
  ls = rs = 1;
  for (auto k : seg) {
    while (ls < k.l) del(ls++);
    while (k.l < ls) add(--ls);
    while (rs <= k.r) add(rs++);
    while (k.r + 1 < rs) del(--rs);
    answer[k.idx] = ans;
  }
  for (int i = 0; i < seg.size(); i++) {
    printf("%d\n", answer[i]);
  }
  return 0;
}
