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;
struct query {
  int l, r, block, i;
  query() { l = r = block = i = -1; }
  query(int _l, int _r, int _i) {
    l = _l;
    r = _r;
    i = _i;
  }
};
query q[100005];
bool cmp(const query& a, const query& b) {
  if (a.block != b.block)
    return a.block < b.block;
  else
    return a.r < b.r;
}
int n, m, sq;
int cnt[100005];
int64_t a[100005];
int answer[100005], ans;
void add(int x) {
  if (cnt[x] == x) ans--;
  cnt[x]++;
  if (cnt[x] == x) ans++;
}
void del(int x) {
  if (cnt[x] == x) ans--;
  cnt[x]--;
  if (cnt[x] == x) ans++;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] >= 100005) a[i] = -1;
  }
  sq = sqrt(n);
  for (int i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    q[i] = query(l - 1, r - 1, i);
    q[i].block = (l - 1) / sq;
  }
  sort(q, q + m, cmp);
  int l = q[0].l, r = q[0].r;
  for (int i = l; i <= r; i++) add(a[i]);
  for (int i = 0; i < m; i++) {
    while (l < q[i].l) del(a[l++]);
    while (l > q[i].l) add(a[--l]);
    while (r < q[i].r) add(a[++r]);
    while (r > q[i].r) del(a[r--]);
    answer[q[i].i] = ans;
  }
  for (int i = 0; i < m; i++) cout << answer[i] << "\n";
  return 0;
}
