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];
vector<pair<int, pair<int, int> > > q;
int cnt[100010];
int ans = 0;
int ret[100010];
int n;
bool com(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
  if (a.first / 555 != b.first / 555) {
    return a.first / 555 < b.first / 555;
  }
  return a.second.first < b.second.first;
}
void add(int ind) {
  if (a[ind] > n) return;
  cnt[a[ind]]++;
  if (cnt[a[ind]] == a[ind])
    ans++;
  else if (cnt[a[ind]] == a[ind] + 1)
    ans--;
}
void rem(int ind) {
  if (a[ind] > n) return;
  cnt[a[ind]]--;
  if (cnt[a[ind]] == a[ind])
    ans++;
  else if (cnt[a[ind]] == a[ind] - 1)
    ans--;
}
int main(void) {
  int m;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  for (int i = 0; i < m; ++i) {
    int l, r;
    scanf("%d %d", &l, &r);
    --l, --r;
    q.push_back(make_pair(l, make_pair(r, i)));
  }
  sort(q.begin(), q.end(), com);
  int currl = 0, currr = 0;
  for (int i = 0; i < m; ++i) {
    int l = q[i].first;
    int r = q[i].second.first;
    int qq = q[i].second.second;
    while (currl < l) {
      rem(currl);
      currl++;
    }
    while (currl > l) {
      add(currl - 1);
      currl--;
    }
    while (currr <= r) {
      add(currr);
      currr++;
    }
    while (currr > r + 1) {
      rem(currr - 1);
      currr--;
    }
    ret[qq] = ans;
  }
  for (int i = 0; i < m; ++i) printf("%d\n", ret[i]);
  return 0;
}
