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;
void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long n, q, bs;
long long input[2000000];
long long ans[2000000];
unordered_map<long long, long long> cnt;
struct query {
  long long L, R, indx;
} arr[2000000];
bool cmp(const query &n1, const query &n2) {
  long long block1 = n1.L / bs;
  long long block2 = n2.L / bs;
  if (block1 == block2)
    return n1.R < n2.R;
  else
    return block1 < block2;
}
long long sum = 0;
void add(long long i) {
  cnt[input[i]]++;
  if (cnt[input[i]] == input[i]) sum++;
  if (cnt[input[i]] == input[i] + 1) sum--;
}
void del(long long i) {
  cnt[input[i]]--;
  if (cnt[input[i]] == input[i]) sum++;
  if (cnt[input[i]] == input[i] - 1) sum--;
}
int main() {
  scanf("%lld %lld", &n, &q);
  bs = sqrt(n);
  for (long long i = 0; i < n; i++) {
    scanf("%lld", &input[i]);
  }
  for (long long i = 0; i < q; i++) {
    scanf("%lld %lld", &arr[i].L, &arr[i].R);
    --arr[i].L;
    --arr[i].R;
    arr[i].indx = i;
  }
  sort(arr, arr + q, cmp);
  long long left = 0, right = -1;
  sum = 0;
  for (long long i = 0; i < q; i++) {
    long long l = arr[i].L;
    long long r = arr[i].R;
    long long in = arr[i].indx;
    while (l > left) {
      del(left++);
    }
    while (l < left) {
      add(--left);
    }
    while (r > right) {
      add(++right);
    }
    while (r < right) {
      del(right--);
    }
    ans[in] = sum;
  }
  for (long long i = 0; i < q; i++) printf("%lld\n", ans[i]);
  return 0;
}
