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 = 337, i, n, m, a[100505], aa, bb;
vector<pair<int, pair<int, int> > > v;
int mark[100505];
vector<int> vv;
bool cmp(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
  if (a.first / block != b.first / block)
    return a.first / block < b.first / block;
  return a.second < b.second;
}
int L, R, val, ans[100505];
void add(int x) {
  if (x > n) return;
  mark[x]++;
  if (mark[x] == x) val++;
  if (mark[x] == x + 1) val--;
}
void rem(int x) {
  if (x > n) return;
  mark[x]--;
  if (mark[x] == x - 1) val--;
  if (mark[x] == x) val++;
}
int main() {
  cin >> n >> m;
  for (i = 0; i < n; i++) cin >> a[i];
  for (i = 0; i < m; i++) {
    cin >> aa >> bb;
    aa--;
    bb--;
    v.push_back(make_pair(aa, make_pair(bb, i)));
  }
  sort(v.begin(), v.end(), cmp);
  L = R = 0;
  val = 0;
  for (i = 0; i < v.size(); i++) {
    int index = v[i].second.second;
    while (L < v[i].first) {
      rem(a[L]);
      L++;
    }
    while (L > v[i].first) {
      L--;
      add(a[L]);
    }
    while (R <= v[i].second.first) {
      add(a[R]);
      R++;
    }
    while (R - 1 >= 0 && R - 1 > v[i].second.first) {
      rem(a[R - 1]);
      R--;
    }
    ans[index] = val;
  }
  for (i = 0; i < m; i++) {
    cout << ans[i] << "\n";
  }
  return 0;
}
