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 N = 1e5 + 5;
int n, m, cnt = 0, sqn;
int d[N];
int ans[N];
int mp[N];
struct node {
  int id;
  int l;
  int r;
  bool friend operator<(node a, node b) {
    if (((a.l - 1) / sqn + 1) == ((b.l - 1) / sqn + 1)) return a.r < b.r;
    return ((a.l - 1) / sqn + 1) < ((b.l - 1) / sqn + 1);
  }
} p[N];
void add(int pos) {
  pos = d[pos];
  if (pos > 100000) return;
  mp[pos]++;
  if (mp[pos] == pos) cnt++;
  if (mp[pos] == pos + 1) cnt--;
}
void del(int pos) {
  pos = d[pos];
  if (pos > 100000) return;
  if (mp[pos] == pos) cnt--;
  if (mp[pos] == pos + 1) cnt++;
  mp[pos]--;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &d[i]);
  }
  sqn = (int)sqrt(n);
  for (int i = 1; i <= m; i++) {
    p[i].id = i;
    scanf("%d %d", &p[i].l, &p[i].r);
  }
  sort(p + 1, p + 1 + m);
  int x = 1, y = 0;
  for (int i = 1; i <= m; i++) {
    int u = p[i].l;
    int v = p[i].r;
    while (x > u) x--, add(x);
    while (y < v) y++, add(y);
    while (x < u) del(x), x++;
    while (y > v) del(y), y--;
    ans[p[i].id] = cnt;
  }
  for (int i = 1; i <= m; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
