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;
bool __insideGrid(int x, int y, int h, int w, int sx = 0, int sy = 0) {
  if (x < sx) return false;
  if (y < sy) return false;
  if (x >= sx + h) return false;
  if (y >= sy + w) return false;
  return true;
}
const double pi = acos(-1.0);
int data[100005];
int freq[100005];
bool visited[100005] = {0};
int occur[500][100005];
int num[500];
int main() {
  int n, m;
  scanf("%d", &(n));
  scanf("%d", &(m));
  for (int i = 1; i <= n; i++) {
    scanf("%d", &(data[i]));
    if (data[i] <= n) freq[data[i]]++;
  }
  int occurSZ = 0;
  for (int i = 1; i <= n; i++) {
    if (data[i] > n) continue;
    if (visited[data[i]]) continue;
    if (freq[data[i]] >= data[i]) {
      visited[data[i]] = 1;
      num[occurSZ] = data[i];
      occur[occurSZ][0] = 0;
      for (int j = 1; j <= n; j++) {
        if (data[j] == data[i])
          occur[occurSZ][j] = occur[occurSZ][j - 1] + 1;
        else
          occur[occurSZ][j] = occur[occurSZ][j - 1];
      }
      occurSZ++;
    }
  }
  int L, R;
  for (int i = 0; i < m; i++) {
    scanf("%d", &(L));
    scanf("%d", &(R));
    int ans = 0;
    for (int j = 0; j < occurSZ; j++) {
      int temp = occur[j][R] - occur[j][L - 1];
      if (temp == num[j]) ans++;
    }
    printf("%d\n", ans);
  }
  return 0;
}
