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 MXN = 1e5 + 5, BLOCK_SIZE = 20000;
struct Query {
  int l, r, idx;
  bool operator<(const Query &y) const {
    int x_block = l / BLOCK_SIZE;
    int y_block = y.l / BLOCK_SIZE;
    if (x_block == y_block) return r < y.r;
    return x_block < y_block;
  }
};
int A[MXN], final_ans[MXN], freq[MXN];
int cur_ans;
void operate(int idx, int delta) {
  int x = A[idx];
  if (x >= MXN) return;
  if (freq[x] == x) cur_ans--;
  freq[x] += delta;
  if (freq[x] == x) cur_ans++;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int N, M;
  cin >> N >> M;
  for (int i = 0; i < N; i++) cin >> A[i];
  vector<Query> queries(M);
  for (int i = 0; i < M; i++) {
    cin >> queries[i].l >> queries[i].r;
    queries[i].l--, queries[i].r--;
    queries[i].idx = i;
  }
  sort(queries.begin(), queries.end());
  int i = 0, j = -1;
  for (const auto &[l, r, idx] : queries) {
    while (j < r) operate(++j, 1);
    while (i > l) operate(--i, 1);
    while (i < l) operate(i++, -1);
    while (j > r) operate(j--, -1);
    final_ans[idx] = cur_ans;
  }
  for (int i = 0; i < M; i++) cout << final_ans[i] << "\n";
}
