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 {
  long long l, r, id, b;
} qr[100009];
long long k;
long long ar[100009], anss[100009];
long long ans = 0, l = 1, r = 0;
bool f = false;
;
unordered_map<long long, long long> c;
bool com(query c, query d) {
  if (c.b == d.b) return c.r < d.r;
  return c.b < d.b;
}
void add(int x) {
  long long i = ar[x];
  c[i]++;
  if (c[i] == i) {
    ans++;
  }
  if (c[i] == i + 1) ans--;
}
void remove(int x) {
  long long i = ar[x];
  if (c[i] == i) ans--;
  c[i]--;
  if (c[i] == i) ans++;
}
int main() {
  long long n, t;
  scanf("%lld", &n), scanf("%lld", &t);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &ar[i]);
  }
  k = sqrt(n);
  for (long long i = 0; i < t; i++) {
    scanf("%lld", &qr[i].l), scanf("%lld", &qr[i].r);
    qr[i].id = i;
    qr[i].b = qr[i].l / k;
  }
  sort(qr, qr + t, com);
  for (long long i = 0; i < t; i++) {
    while (l > qr[i].l) {
      l--;
      add(l);
    };
    while (r < qr[i].r) {
      r++;
      add(r);
    };
    while (l < qr[i].l) {
      remove(l);
      l++;
    }
    while (r > qr[i].r) {
      remove(r);
      r--;
    }
    anss[qr[i].id] = ans;
  }
  for (long long i = 0; i < t; i++) printf("%lld", anss[i]), printf("\n");
}
