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 dat[100100];
vector<int> cands;
vector<int> ids[450];
int valids[100100][2];
map<int, int> total;
int l[100100], r[100100], ans[100100];
int main() {
  int N, M;
  scanf("%d%d", &N, &M);
  for (int i = 0; i < N; i++) {
    scanf("%d", dat + i);
    total[dat[i]]++;
  }
  map<int, int>::iterator it = total.begin();
  while (it != total.end()) {
    if ((*it).first <= (*it).second) {
      cands.push_back((*it).first);
    }
    it++;
  }
  for (int i = 0; i < cands.size(); i++) {
    ids[i].push_back(-1);
    for (int j = 0; j < N; j++) {
      if (dat[j] == cands[i]) {
        ids[i].push_back(j);
      }
    }
    ids[i].push_back(N);
  }
  for (int i = 0; i < M; i++) {
    scanf("%d%d", l + i, r + i);
    ans[i] = 0;
    l[i]--;
    r[i]--;
  }
  for (int i = 0; i < cands.size(); i++) {
    int val = cands[i];
    memset(valids, -1, sizeof(valids));
    for (int j = 0; j + 1 + val < ids[i].size(); j++) {
      for (int k = ids[i][j] + 1; k <= ids[i][j + 1]; k++) {
        valids[k][0] = ids[i][j + val];
        valids[k][1] = ids[i][j + val + 1];
      }
    }
    for (int j = 0; j < M; j++) {
      if (valids[l[j]][0] <= r[j] && valids[l[j]][1] > r[j]) {
        ans[j]++;
      }
    }
  }
  for (int i = 0; i < M; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
