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 const N = 1e5 + 5, mod = 1e9 + 7, inf = 1e17, maxn = 500;
int pl[maxn], pr[N][maxn];
long long a[N], b[N], cnt[N], ans[N];
bool mark[N];
vector<pair<int, int> > vt;
vector<pair<pair<int, int>, int> > vec;
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  long long n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  b[n] = 0;
  long long f = 1, k = 0;
  for (int i = 1; i <= n; i++) {
    if (b[i] == b[i - 1]) {
      f++;
    } else {
      if (f >= b[i - 1]) {
        vt.push_back({b[i - 1], f - b[i - 1]});
        cnt[b[i - 1]] = k + 1;
        k++;
      }
      if (b[i] > n) {
        break;
      }
      f = 1;
    }
  }
  for (int i = 0; i < n; i++) {
    if (i != 0) {
      for (int j = 0; j < k; j++) {
        pr[n - i - 1][j] = pr[n - i][j];
      }
    }
    if (a[n - i - 1] <= n && cnt[a[n - i - 1]]) {
      pr[n - i - 1][cnt[a[n - i - 1]] - 1]++;
    }
  }
  for (int i = 0; i < m; i++) {
    long long l, r;
    cin >> l >> r;
    l -= 2;
    vec.push_back({{l, r}, i});
  }
  long long q = 0;
  sort(vec.begin(), vec.end());
  for (int i = 0; i < m; i++) {
    while (q <= vec[i].first.first) {
      if (a[q] <= n && cnt[a[q]]) {
        pl[cnt[a[q]] - 1]++;
      }
      q++;
    }
    long long it = 0;
    for (auto x : vt) {
      long long a1 = pl[it];
      long long a2 = pr[vec[i].first.second][it];
      if ((a1 + a2) == x.second) {
        ans[vec[i].second]++;
      }
      it++;
    }
  }
  for (int i = 0; i < m; i++) {
    cout << ans[i] << endl;
  }
}
