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;
const int maxn = 1e5 + 5, mod = 1e9 + 7;
struct Data {
  int l, r, id;
  Data() {}
  Data(int l, int r, int id) : l(l), r(r), id(id) {}
};
int n, q, L, ans[maxn], a[maxn], mp[maxn];
Data d[maxn];
bool block(int first, int second) { return first / L == second / L; }
bool cmp(Data first, Data second) {
  if (!block(first.l, second.l))
    return first.l / L < second.l / L;
  else
    return first.r < second.r;
}
void process(int l, int r, int &res, int val) {
  for (int i = l; i <= r; ++i)
    if (a[i] <= n) {
      int &temp = mp[a[i]];
      if (temp == a[i]) --res;
      temp += val;
      if (temp == a[i]) ++res;
    }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  L = sqrt(n);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i <= q; ++i) {
    int l, r;
    cin >> l >> r;
    d[i] = Data(l, r, i);
  }
  sort(d + 1, d + q + 1, cmp);
  process(d[1].l, d[1].r, ans[d[1].id], 1);
  for (int i = 2; i <= q; ++i) {
    int res = ans[d[i - 1].id], l = d[i - 1].l, r = d[i - 1].r;
    if (l < d[i].l)
      process(l, d[i].l - 1, res, -1);
    else
      process(d[i].l, l - 1, res, 1);
    if (r < d[i].r)
      process(r + 1, d[i].r, res, 1);
    else
      process(d[i].r + 1, r, res, -1);
    ans[d[i].id] = res;
  }
  for (int i = 1; i <= q; ++i) cout << ans[i] << '\n';
  return 0;
}
