Cosider a sequence, consisting of n integers: a1, a2, ..., an. Jeff can perform the following operation on sequence a:

  * take three integers v, t, k (1 ≤ v, t ≤ n; 0 ≤ k; v + tk ≤ n), such that av = av + t, av + t = av + 2t, ..., av + t(k - 1) = av + tk; 
  * remove elements av, av + t, ..., av + t·k from the sequence a, the remaining elements should be reindexed a1, a2, ..., an - k - 1. 
  * permute in some order the remaining elements of sequence a. 



A beauty of a sequence a is the minimum number of operations that is needed to delete all elements from sequence a.

Jeff's written down a sequence of m integers b1, b2, ..., bm. Now he wants to ask q questions. Each question can be described with two integers li, ri. The answer to the question is the beauty of sequence bli, bli + 1, ..., bri. You are given the sequence b and all questions. Help Jeff, answer all his questions.

Input

The first line contains integer m (1 ≤ m ≤ 105). The next line contains m integers b1, b2, ..., bm (1 ≤ bi ≤ 105). 

The third line contains integer q (1 ≤ q ≤ 105) — the number of questions. The next q lines contain pairs of integers, i-th of them contains a pair of integers li, ri (1 ≤ li ≤ ri ≤ m) — the description of i-th question.

Output

In q lines print the answers to Jeff's queries. Print the answers according to the order of questions in input.

Examples

Input

5
2 2 1 1 2
5
1 5
1 1
2 2
1 3
2 3


Output

2
1
1
2
2


Input

10
2 1 3 3 3 3 1 3 1 1
10
4 8
2 10
1 10
4 4
1 3
2 4
6 7
1 9
2 5
1 1


Output

2
3
3
1
3
2
2
3
2
1
#include <bits/stdc++.h>
using namespace std;
int a[200005];
struct val {
  int l;
  int r;
  int i;
} q[200005];
int sq;
int mp[200005];
deque<int> s[200005];
deque<int> w[200005];
int ok[200005];
int ans[200005];
bool comp(val a, val b) {
  if (a.l / sq != b.l / sq)
    return a.l / sq < b.l / sq;
  else
    return a.r < b.r;
}
int main() {
  int n;
  scanf("%d", &n);
  sq = sqrt(n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  int qu;
  scanf("%d", &qu);
  int l, r;
  for (int i = 1; i <= qu; i++) {
    scanf("%d %d", &l, &r);
    q[i].l = l;
    q[i].r = r;
    q[i].i = i;
  }
  sort(q + 1, q + qu + 1, comp);
  l = q[1].l;
  r = l - 1;
  int cnt = 0;
  int all = 0;
  for (int i = 1; i <= qu; i++) {
    while (l < q[i].l) {
      if (r >= l) {
        mp[a[l]]--;
        if (mp[a[l]] == 0) cnt--;
        all -= (ok[a[l]] == 0) && !(s[a[l]].empty());
        s[a[l]].pop_front();
        if (s[a[l]].empty() != 1) {
          w[a[l]].pop_front();
          if (w[a[l]].empty() != 1)
            ok[a[l]] -= (s[a[l]].front() - l) != w[a[l]].front();
        }
        all += (ok[a[l]] == 0) && !(s[a[l]].empty());
      }
      l++;
    }
    while (l > q[i].l) {
      l--;
      if (r >= l) {
        mp[a[l]]++;
        if (mp[a[l]] == 1) cnt++;
        all -= (ok[a[l]] == 0) && !(s[a[l]].empty());
        if (!s[a[l]].empty()) {
          if (!w[a[l]].empty())
            ok[a[l]] += (s[a[l]].front() - l) != w[a[l]].front();
          w[a[l]].push_front(s[a[l]].front() - l);
        }
        s[a[l]].push_front(l);
        all += (ok[a[l]] == 0) && !(s[a[l]].empty());
      }
    }
    while (r < q[i].r) {
      r++;
      if (r >= l) {
        mp[a[r]]++;
        if (mp[a[r]] == 1) cnt++;
        all -= (ok[a[r]] == 0) && !(s[a[r]].empty());
        if (!(s[a[r]].empty())) {
          if (!w[a[r]].empty())
            ok[a[r]] += (r - s[a[r]].back()) != w[a[r]].back();
          w[a[r]].push_back(r - s[a[r]].back());
        }
        s[a[r]].push_back(r);
        all += (ok[a[r]] == 0) && !(s[a[r]].empty());
      }
    }
    while (r > q[i].r) {
      if (r >= l) {
        mp[a[r]]--;
        if (mp[a[r]] == 0) cnt--;
        all -= (ok[a[r]] == 0) && (!s[a[r]].empty());
        s[a[r]].pop_back();
        if (!s[a[r]].empty()) {
          w[a[r]].pop_back();
          if (!w[a[r]].empty())
            ok[a[r]] -= (r - s[a[r]].back()) != w[a[r]].back();
        }
        all += (ok[a[r]] == 0) && !(s[a[r]].empty());
      }
      r--;
    }
    ans[q[i].i] = cnt + (all == 0);
  }
  for (int i = 1; i <= qu; i++) cout << ans[i] << "\n";
  return 0;
}
