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 g;
int A[100100];
int cnt[100100];
vector<pair<int, pair<int, int>>> Q;
vector<pair<int, int>> R;
int a;
bool compare(pair<int, pair<int, int>> a, pair<int, pair<int, int>> b) {
  if (a.second.first / g == b.second.first / g) {
    return a.second.second < b.second.second;
  } else {
    return a.second.first / g < b.second.first / g;
  }
}
void add(int v) {
  if (v > 100000) return;
  if (cnt[v] == v) a--;
  cnt[v]++;
  if (cnt[v] == v) a++;
}
void remove(int v) {
  if (v > 100000) return;
  if (cnt[v] == v) a--;
  cnt[v]--;
  if (cnt[v] == v) a++;
}
int main() {
  int n, m;
  scanf(" %d %d", &n, &m);
  g = sqrt(n);
  a = 0;
  for (int i = 0; i < n; i++) {
    scanf(" %d", &A[i]);
  }
  for (int i = 0; i < m; i++) {
    int l, r;
    scanf(" %d %d", &l, &r);
    Q.push_back(make_pair(i, make_pair(l - 1, r - 1)));
  }
  sort(Q.begin(), Q.end(), compare);
  int l = -1, r = -1;
  a = 0;
  for (int i = 0; i < m; i++) {
    while (l < Q[i].second.first) {
      if (l != -1) {
        remove(A[l]);
      }
      l++;
    }
    while (l > Q[i].second.first) add(A[--l]);
    while (r < Q[i].second.second) add(A[++r]);
    while (r > Q[i].second.second) remove(A[r--]);
    R.push_back(make_pair(Q[i].first, a));
  }
  sort(R.begin(), R.end());
  for (int i = 0; i < m; i++) {
    printf("%d\n", R[i].second);
  }
  return 0;
}
