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 rev[100005], ar[100005], track[100005], cnt, ans[100005];
vector<pair<pair<int, int>, pair<int, int> > > v;
map<int, int> save;
void reduce(int val, int actual) {
  track[val]--;
  if (track[val] == actual - 1)
    cnt--;
  else if (track[val] == actual)
    cnt++;
}
void add(int val, int actual) {
  track[val]++;
  if (track[val] == actual)
    cnt++;
  else if (track[val] == actual + 1)
    cnt--;
}
int main() {
  int tc, t, n, q, i, j, l, r;
  set<int> st;
  scanf("%d %d", &n, &q);
  for (i = 1; i <= n; i++) {
    scanf("%d", &ar[i]);
    st.insert(ar[i]);
  }
  i = 1;
  for (auto it = st.begin(); it != st.end(); it++, i++) {
    save[*it] = i;
    rev[i] = *it;
  }
  for (i = 1; i <= n; i++) {
    ar[i] = save[ar[i]];
  }
  int blk = sqrt(n);
  for (i = 0; i < q; i++) {
    scanf("%d%d", &l, &r);
    v.push_back(make_pair(make_pair(l / blk, r), make_pair(l, i)));
  }
  sort(v.begin(), v.end());
  int lp = 0, rp = 0;
  for (i = 0; i < q; i++) {
    l = v[i].second.first;
    r = v[i].first.second;
    while (lp < l) {
      int val = ar[lp];
      lp++;
      reduce(val, rev[val]);
    }
    while (lp > l) {
      lp--;
      int val = ar[lp];
      add(val, rev[val]);
    }
    while (rp <= r) {
      int val = ar[rp];
      rp++;
      add(val, rev[val]);
    }
    while (rp > r + 1) {
      rp--;
      int val = ar[rp];
      reduce(val, rev[val]);
    }
    ans[v[i].second.second] = cnt;
  }
  for (i = 0; i < q; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
