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 mod = (long long)1e9 + 7;
long long n, m, a[100005];
long long sq;
bool comp(pair<pair<long long, long long>, long long> x,
          pair<pair<long long, long long>, long long> y) {
  if ((x.first.first) / sq == (y.first.first) / sq) {
    return x.first.second < y.first.second;
  }
  return (x.first.first) / sq < (y.first.first) / sq;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  for (long long i = 0; i <= n - 1; ++i) {
    cin >> a[i];
    if (a[i] > 100000) a[i] = 100001;
  }
  sq = sqrt(n);
  long long r1, r2;
  pair<pair<long long, long long>, long long> query[m];
  for (long long i = 0; i <= m - 1; ++i) {
    cin >> r1 >> r2;
    --r1, --r2;
    query[i] = make_pair(make_pair(r1, r2), i);
  }
  sort(query, query + m, comp);
  long long ans[100005];
  long long result = 0, begin = query[0].first.first,
            end = query[0].first.second, cnt[100005];
  memset(cnt, 0, sizeof(cnt));
  for (long long i = begin; i <= end; ++i) {
    ++cnt[a[i]];
    if (cnt[a[i]] == a[i] + 1) {
      --result;
    } else if (cnt[a[i]] == a[i]) {
      ++result;
    }
  }
  ans[query[0].second] = result;
  for (long long i = 1; i <= m - 1; ++i) {
    r1 = query[i].first.first;
    r2 = query[i].first.second;
    if (r1 < begin) {
      while (r1 < begin) {
        ++cnt[a[r1]];
        if (cnt[a[r1]] == a[r1])
          ++result;
        else if (cnt[a[r1]] == a[r1] + 1)
          --result;
        ++r1;
      }
    } else if (r1 > begin) {
      --r1;
      while (r1 >= begin) {
        --cnt[a[r1]];
        if (cnt[a[r1]] == a[r1])
          ++result;
        else if (cnt[a[r1]] == a[r1] - 1)
          --result;
        --r1;
      }
    }
    if (r2 > end) {
      while (r2 > end) {
        ++cnt[a[r2]];
        if (cnt[a[r2]] == a[r2])
          ++result;
        else if (cnt[a[r2]] == a[r2] + 1)
          --result;
        --r2;
      }
    } else if (r2 < end) {
      ++r2;
      while (r2 <= end) {
        --cnt[a[r2]];
        if (cnt[a[r2]] == a[r2])
          ++result;
        else if (cnt[a[r2]] == a[r2] - 1)
          --result;
        ++r2;
      }
    }
    ans[query[i].second] = result;
    begin = query[i].first.first;
    end = query[i].first.second;
  }
  for (long long i = 0; i <= m - 1; ++i) cout << ans[i] << "\n";
  return 0;
}
