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], flag[450], Count[450][100010], N, M, cnt;
map<int, int> mapnum;
int main() {
  scanf("%d%d", &N, &M);
  for (int i = 1; i <= N; i++) {
    scanf("%d", &a[i]);
    if (mapnum.count(a[i]))
      (mapnum.find(a[i])->second)++;
    else
      mapnum.insert(make_pair(a[i], 1));
  }
  map<int, int>::iterator iter;
  cnt = 0;
  for (iter = mapnum.begin(); iter != mapnum.end(); iter++) {
    if (iter->first <= iter->second) flag[cnt++] = iter->first;
  }
  for (int i = 0; i < cnt; i++) {
    Count[i][0] = 0;
    for (int j = 1; j <= N; j++)
      if (a[j] == flag[i])
        Count[i][j] = Count[i][j - 1] + 1;
      else
        Count[i][j] = Count[i][j - 1];
  }
  int l, r, ans;
  for (int i = 0; i < M; i++) {
    scanf("%d%d", &l, &r);
    ans = 0;
    for (int j = 0; j < cnt; j++)
      if (Count[j][r] - Count[j][l - 1] == flag[j]) ans++;
    printf("%d\n", ans);
  }
  return 0;
}
