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;
unordered_map<long long, long long> mp;
struct node {
  long long l, r, id, blc;
};
node q[100005];
long long arr[100005];
long long ans[100005];
long long sum;
void add(long long x) {
  x = arr[x];
  if (mp[x] == x) {
    sum--;
  }
  mp[x]++;
  if (mp[x] == x) {
    sum++;
  }
}
void rem(long long x) {
  x = arr[x];
  if (mp[x] == x) sum--;
  mp[x]--;
  if (mp[x] == x) {
    sum++;
  }
}
bool comp(node a, node b) {
  if (a.blc != b.blc) {
    return a.blc < b.blc;
  }
  return a.r < b.r;
}
int main() {
  long long i, j, k, l, m, n, o;
  long long testcase;
  long long input, flag, tag;
  scanf("%lld", &n);
  scanf("%lld", &m);
  for (i = 1; i <= n; i++) scanf("%lld", &arr[i]);
  long long ssq = sqrt(n) + 5;
  for (i = 1; i <= m; i++) {
    scanf("%lld", &j);
    scanf("%lld", &k);
    q[i].l = j;
    q[i].r = k;
    q[i].id = i;
    q[i].blc = j / ssq;
  }
  sort(q + 1, q + 1 + m, comp);
  l = 1;
  long long r = 0;
  sum = 0;
  for (i = 1; i <= m; i++) {
    while (l > q[i].l) add(--l);
    while (r < q[i].r) add(++r);
    while (l < q[i].l) rem(l++);
    while (r > q[i].r) rem(r--);
    ans[q[i].id] = sum;
  }
  for (i = 1; i <= m; i++) {
    printf("%lld\n", ans[i]);
  }
}
