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;
long long n, q;
const long long M = 1e5 + 500;
long long curans;
long long cnt[M];
long long ans[M];
long long blk_sz;
long long ar[M];
pair<pair<long long, long long>, long long> queries[M];
bool saklo(const pair<pair<long long, long long>, long long> &x,
           const pair<pair<long long, long long>, long long> &y) {
  long long blkx = x.first.first / blk_sz;
  long long blky = y.first.first / blk_sz;
  if (blkx != blky) return blkx < blky;
  return x.first.second < y.first.second;
}
void add(long long x) {
  if (x > 100000) return;
  if (cnt[x] == x) curans -= 1;
  cnt[x] += 1;
  if (cnt[x] == x) curans += 1;
}
void remove(long long x) {
  if (x > 100000) return;
  if (cnt[x] == x) curans -= 1;
  cnt[x] -= 1;
  if (cnt[x] == x) curans += 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  blk_sz = sqrt(n);
  for (long long i = 0; i < n; i++) cin >> ar[i];
  for (long long i = 0; i < q; i++) {
    long long x, y;
    cin >> x >> y;
    x--;
    y--;
    queries[i].first.first = x;
    queries[i].first.second = y;
    queries[i].second = i;
  }
  sort(queries, queries + q, saklo);
  long long mo_left = 0, mo_right = -1;
  for (long long i = 0; i < q; i++) {
    long long left = queries[i].first.first;
    long long right = queries[i].first.second;
    while (mo_right < right) add(ar[++mo_right]);
    while (mo_right > right) remove(ar[mo_right--]);
    while (mo_left < left) remove(ar[mo_left++]);
    while (mo_left > left) add(ar[--mo_left]);
    ans[queries[i].second] = curans;
  }
  for (long long i = 0; i < q; i++) cout << ans[i] << "\n";
  return 0;
}
