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 a[100010], I[100010], L[100010], l[100010], r[100010];
int cmp(int u, int v) {
  if (L[u] != L[v]) return L[u] < L[v];
  return r[u] < r[v];
}
int sum, cnt[100010], ans[100010];
int add(int u) {
  if (a[u] > 100010) return 0;
  if (cnt[a[u]] == a[u]) sum--;
  cnt[a[u]]++;
  if (cnt[a[u]] == a[u]) sum++;
}
int del(int u) {
  if (a[u] > 100010) return 0;
  if (cnt[a[u]] == a[u]) sum--;
  cnt[a[u]]--;
  if (cnt[a[u]] == a[u]) sum++;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  int N = (int)(sqrt(n) + 1);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", l + i, r + i);
    l[i]--, r[i]--;
    L[i] = l[i] / N;
    I[i] = i;
  }
  sort(I, I + m, cmp);
  int k = 0;
  for (int i = 0; i <= N; i++) {
    int pre = i * N, cur = pre - 1;
    memset(cnt, 0, sizeof(cnt));
    sum = 0;
    for (int j = i; j <= N; j++) {
      while (k < m && l[I[k]] / N == i && r[I[k]] / N == j) {
        while (l[I[k]] < pre) add(--pre);
        while (r[I[k]] > cur) add(++cur);
        while (l[I[k]] > pre) del(pre++);
        while (r[I[k]] < cur) del(cur--);
        ans[I[k]] = sum;
        k++;
      }
    }
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
  return 0;
}
