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 main() {
  int N, Query;
  scanf("%d%d", &N, &Query);
  static int A[100000], Count[100001];
  memset(Count, 0, sizeof(Count));
  for (int i = 0; i < N; i++) {
    scanf("%d", &A[i]);
    if (A[i] <= N) Count[A[i]]++;
  }
  int M = 0;
  static int All[100000];
  for (int i = 1; i <= N; i++)
    if (Count[i] >= i) All[M++] = i;
  static int Sum[446][100001];
  for (int i = 0; i < M; i++)
    for (int j = Sum[i][0] = 0; j < N; j++)
      Sum[i][j + 1] = Sum[i][j] + (A[j] == All[i]);
  while (Query--) {
    int L, R, Ans = 0;
    scanf("%d%d", &L, &R);
    for (int i = 0; i < M; i++)
      if (Sum[i][R] - Sum[i][L - 1] == All[i]) Ans++;
    printf("%d\n", Ans);
  }
  return 0;
}
