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;
struct dhjk {
  int L, R, id;
};
dhjk qu[100010];
int cnt[100010];
int res, sqn;
int a[100010], n, m, ans[100010];
void remove(int x) {
  if (a[x] > n || a[x] == -1) return;
  if (cnt[a[x]] == a[x]) res--;
  cnt[a[x]]--;
  if (cnt[a[x]] == a[x]) res++;
}
void add(int x) {
  if (a[x] > n || a[x] == -1) return;
  if (cnt[a[x]] == a[x]) res--;
  cnt[a[x]]++;
  if (cnt[a[x]] == a[x]) res++;
}
bool cmp(dhjk x, dhjk y) {
  return (x.L / sqn < y.L / sqn) ||
         (x.L / sqn == y.L / sqn && x.R / sqn < y.R / sqn);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  sqn = sqrt(n);
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= m; ++i) {
    int l, r;
    cin >> l >> r;
    qu[i].L = l;
    qu[i].R = r;
    qu[i].id = i;
  }
  sort(qu + 1, qu + m + 1, cmp);
  int L = 0, R = 0;
  a[0] = -1;
  for (int i = 1; i <= m; ++i) {
    while (L < qu[i].L) {
      remove(L);
      L++;
    }
    while (L > qu[i].L) {
      L--;
      add(L);
    }
    while (R > qu[i].R) {
      remove(R);
      R--;
    }
    while (R < qu[i].R) {
      R++;
      add(R);
    }
    ans[qu[i].id] = res;
  }
  for (int i = 1; i <= m; ++i) cout << ans[i] << endl;
}
