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 MOD = 1e9 + 7;
const int inf = INT_MAX;
const int MAXN = 100100;
const int MAXC = 501;
int N, M;
int num[MAXN];
int numoc[MAXN];
int ps[MAXC][MAXN];
vector<int> poss;
int main() {
  for (int i = 0; i < MAXN; i++) numoc[i] = 0;
  cin >> N >> M;
  for (int i = 0; i < N; i++) {
    cin >> num[i];
    if (num[i] <= N) numoc[num[i]]++;
  }
  for (int i = 1; i <= N; i++)
    if (numoc[i] >= i) {
      poss.push_back(i);
      int loc = poss.size() - 1;
      ps[loc][0] = 0;
      for (int j = 0; j < N; j++) {
        ps[loc][j + 1] = ps[loc][j];
        if (num[j] == i) ps[loc][j + 1]++;
      }
    }
  for (int j = 0; j < M; j++) {
    int ans = 0;
    int l, r;
    cin >> l >> r;
    for (int i = 0; i < poss.size(); i++) {
      if (ps[i][r] - ps[i][l - 1] == poss[i]) ans++;
    }
    cout << ans << "\n";
  }
  return 0;
}
