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 N = 1e5 + 9;
int n, q, len, a[N], ans[N];
int res, cnt[N];
bool good[N];
struct query {
  int l, r, idx;
};
vector<query> v;
bool cmp(query &a, query &b) {
  if (a.l / len != b.l / len) {
    return a.l < b.l;
  }
  return a.r < b.r;
}
void add(int x) {
  if (a[x] > n) return;
  cnt[a[x]]++;
  if (cnt[a[x]] == a[x]) {
    res++;
    good[a[x]] = 1;
  } else {
    if (good[a[x]]) {
      res--;
      good[a[x]] = 0;
    }
  }
}
void rmv(int x) {
  if (a[x] > n) return;
  if (cnt[a[x]] == a[x]) {
    res--;
    good[a[x]] = 0;
  }
  cnt[a[x]]--;
  if (cnt[a[x]] == a[x]) {
    res++;
    good[a[x]] = 1;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  len = sqrt(n) + 1;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int l, r, i = 0; i < q; i++) {
    cin >> l >> r;
    l--;
    r--;
    query tmp;
    tmp.l = l;
    tmp.r = r;
    tmp.idx = i;
    v.push_back(tmp);
  }
  int l = 0, r = -1;
  sort(v.begin(), v.end(), cmp);
  for (int i = 0; i < q; i++) {
    while (r < v[i].r) {
      r++;
      add(r);
    }
    while (r > v[i].r) {
      rmv(r);
      r--;
    }
    while (l < v[i].l) {
      rmv(l);
      l++;
    }
    while (l > v[i].l) {
      l--;
      add(l);
    }
    ans[v[i].idx] = res;
  }
  for (int i = 0; i < q; i++) {
    cout << ans[i] << '\n';
  }
}
