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;
int times[100005];
int sum;
int ans[100005];
int num[100005];
int pos[100005];
struct q {
  int l;
  int r;
  int id;
} a[100005];
int n, m;
void init() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &num[i]);
  int block = int(sqrt(n));
  for (int i = 1; i <= n; i++) pos[i] = (i - 1) / block + 1;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a[i].l, &a[i].r);
    a[i].id = i;
  }
}
bool cmp(q a, q b) {
  if (pos[a.l] == pos[b.l]) return a.r < b.r;
  return pos[a.l] < pos[b.l];
}
void upd(int x, int op) {
  if (num[x] > 100000) return;
  if (times[num[x]] == num[x]) sum--;
  if (op == 1)
    times[num[x]]++;
  else
    times[num[x]]--;
  if (times[num[x]] == num[x]) sum++;
}
int main() {
  init();
  sort(a + 1, a + 1 + m, cmp);
  sum = 0;
  for (int i = 1; i <= n; i++) upd(i, 1);
  for (int i = 1, l = 1, r = n; i <= m; i++) {
    while (l < a[i].l) upd(l++, 0);
    while (l > a[i].l) upd(--l, 1);
    while (r < a[i].r) upd(++r, 1);
    while (r > a[i].r) upd(r--, 0);
    ans[a[i].id] = sum;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
}
