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], ret[100010], SIZE, ans;
int cnt[100010];
pair<pair<int, int>, int> qu[100010];
inline bool comp(const pair<pair<int, int>, int> &x,
                 const pair<pair<int, int>, int> &y) {
  int bx = x.first.first / SIZE;
  int by = y.first.first / SIZE;
  if (bx != by) return bx < by;
  return x.first.second < y.first.second;
}
inline void add(int x) {
  if (x != 0) {
    if (cnt[x] == x) ans--;
    cnt[x]++;
    if (cnt[x] == x) ans++;
  }
}
inline void rm(int x) {
  if (x != 0) {
    if (cnt[x] == x) ans--;
    cnt[x]--;
    if (cnt[x] == x) ans++;
  }
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  SIZE = (int)sqrt(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] > n) a[i] = 0;
  }
  for (int i = 0; i < q; i++) {
    scanf("%d %d", &qu[i].first.first, &qu[i].first.second);
    qu[i].first.first--, qu[i].first.second--;
    qu[i].second = i;
  }
  sort(qu, qu + q, comp);
  int L = 0, R = -1;
  ans = 0;
  for (int i = 0; i < q; i++) {
    int left = qu[i].first.first;
    int right = qu[i].first.second;
    while (R < right) {
      R++;
      add(a[R]);
    }
    while (R > right) {
      rm(a[R]);
      R--;
    }
    while (L < left) {
      rm(a[L]);
      L++;
    }
    while (L > left) {
      L--;
      add(a[L]);
    }
    ret[qu[i].second] = ans;
  }
  for (int i = 0; i < q; i++) printf("%d\n", ret[i]);
  return 0;
}
