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 n, m;
struct Q {
  long long l, r, id;
  long long t;
  bool operator<(const Q &b) const {
    if (t == b.t)
      return r < b.r;
    else
      return t < b.t;
  }
} q[100005];
long long Ans = 0;
unordered_map<long long, long long> cnt;
long long num[100005];
long long san[100005];
long long tot = 0;
void add(long long x) {
  if (cnt[num[x]] == num[x]) {
    Ans--;
  } else if (cnt[num[x]] == num[x] - 1)
    Ans++;
  cnt[num[x]]++;
}
void del(long long x) {
  if (cnt[num[x]] == num[x]) {
    Ans--;
  } else if (cnt[num[x]] == num[x] + 1)
    Ans++;
  cnt[num[x]]--;
}
long long ans[100005];
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &num[i]);
  }
  int len = sqrt(n);
  for (int i = 1; i <= m; i++) {
    scanf("%I64d%I64d", &q[i].l, &q[i].r);
    q[i].id = i;
    q[i].t = (q[i].l - 1) / len + 1;
  }
  sort(q + 1, q + 1 + m);
  int l = 1, r = 0;
  for (int i = 1; i <= m; i++) {
    while (l > q[i].l) add(l - 1), l--;
    while (l < q[i].l) del(l), l++;
    while (r < q[i].r) add(r + 1), r++;
    while (r > q[i].r) del(r), r--;
    ans[q[i].id] = Ans;
  }
  for (int i = 1; i <= m; i++) {
    printf("%I64d\n", ans[i]);
  }
  return 0;
}
