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 node {
  int l, r, i;
} ask[100002];
int sum;
int ans[100002];
int v[100002];
int n, m, num;
int b[100002];
int t[100002];
bool cmp(node x, node y) {
  if (b[x.l] == b[y.l]) {
    return x.r < y.r;
  }
  return x.l < y.l;
}
void add(int x) {
  if (v[x] > n) return;
  if (t[v[x]] == v[x]) sum--;
  t[v[x]]++;
  if (t[v[x]] == v[x]) sum++;
}
void del(int x) {
  if (v[x] > n) return;
  if (t[v[x]] == v[x]) sum--;
  t[v[x]]--;
  if (t[v[x]] == v[x]) sum++;
}
int main() {
  cin >> n >> m;
  num = sqrt(n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &v[i]);
    b[i] = (i + num - 1) / num;
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &ask[i].l, &ask[i].r);
    ask[i].i = i;
  }
  sort(ask + 1, ask + m + 1, cmp);
  int l = 1, r = 0;
  for (int i = 1; i <= m; i++) {
    while (r < ask[i].r) add(++r);
    while (r > ask[i].r) del(r--);
    while (l < ask[i].l) del(l++);
    while (l > ask[i].l) add(--l);
    ans[ask[i].i] = sum;
  }
  for (int i = 1; i <= m; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
