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;
int block, cnt[maxn + 5], arr[maxn + 5], L, R, res[maxn + 5], n, m, tot;
struct query {
  bool operator<(const query& rhs) const {
    return (L / block) == (rhs.L / block) ? R < rhs.R : L < rhs.L;
  }
  int L = 0, R = 0, id = 0;
} q[maxn + 5];
inline void Add(int index) {
  int v = arr[index];
  if (v > n) return;
  if (cnt[v] == v) --tot;
  ++cnt[v];
  if (cnt[v] == v) ++tot;
}
inline void Del(int index) {
  int v = arr[index];
  if (v > n) return;
  if (cnt[v] == v) --tot;
  --cnt[v];
  if (cnt[v] == v) ++tot;
}
int main(void) {
  scanf("%d %d", &n, &m);
  block = sqrt(n) + 1;
  for (int i = 1; i <= n; ++i) scanf("%d", &arr[i]);
  for (int i = 1; i <= m; ++i) {
    scanf("%d %d", &q[i].L, &q[i].R);
    q[i].id = i;
  }
  sort(q + 1, q + 1 + m);
  L = 1;
  for (int i = 1; i <= m; ++i) {
    while (R < q[i].R) Add(++R);
    while (R > q[i].R) Del(R--);
    while (L < q[i].L) Del(L++);
    while (L > q[i].L) Add(--L);
    res[q[i].id] = tot;
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", res[i]);
  return 0;
}
