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;
const int MAXN = 1e5 + 10;
const int SQRT = 330;
int n, q, a[MAXN], ans[MAXN], dd[MAXN], td[MAXN], prv[MAXN], last[MAXN],
    cnt[MAXN];
pair<pair<int, int>, int> vec[MAXN];
bool mark[MAXN];
deque<int> qq[MAXN];
bool cmp(pair<pair<int, int>, int> u, pair<pair<int, int>, int> v) {
  if (u.first.first / SQRT != v.first.first / SQRT)
    return u.first.first < v.first.first;
  return u.first.second < v.first.second;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  memset(last, -1, sizeof(last));
  for (int i = 0; i < n; i++) prv[i] = last[a[i]], last[a[i]] = i;
  for (int i = 0; i < n; i++) {
    if (prv[i] == -1)
      dd[i] = -1;
    else {
      if (prv[prv[i]] == -1 || i - prv[i] == prv[i] - prv[prv[i]])
        dd[i] = dd[prv[i]];
      else
        dd[i] = prv[prv[i]];
    }
  }
  cin >> q;
  for (int i = 0; i < q; i++)
    cin >> vec[i].first.first >> vec[i].first.second, vec[i].first.first--,
        vec[i].second = i;
  sort(vec, vec + q, cmp);
  int l = 0, r = 0, cc = 0, cn = 0;
  for (int i = 0; i < q; i++) {
    int tl = vec[i].first.first, tr = vec[i].first.second;
    while (r < tr) {
      if (cnt[a[r]]++ == 0) cc++;
      if (mark[a[r]]) cn--, mark[a[r]] = 0;
      if (dd[r] < l) mark[a[r]] = 1, cn++;
      qq[a[r]].push_back(r), r++;
    }
    while (l > tl) {
      l--;
      if (cnt[a[l]]++ == 0) cc++;
      if (mark[a[l]]) cn--, mark[a[l]] = 0;
      qq[a[l]].push_front(l);
      if (dd[qq[a[l]].back()] < l) mark[a[l]] = 1, cn++;
    }
    while (tr < r) {
      r--;
      if (--cnt[a[r]] == 0) cc--;
      if (mark[a[r]]) mark[a[r]] = 0, cn--;
      qq[a[r]].pop_back();
      if (qq[a[r]].size() && dd[qq[a[r]].back()] < l) mark[a[r]] = 1, cn++;
    }
    while (l < tl) {
      if (--cnt[a[l]] == 0) cc--;
      if (mark[a[l]]) mark[a[l]] = 0, cn--;
      qq[a[l]].pop_front();
      if (qq[a[l]].size() && dd[qq[a[l]].back()] <= l) mark[a[l]] = 1, cn++;
      l++;
    }
    if (cn)
      ans[vec[i].second] = cc;
    else
      ans[vec[i].second] = cc + 1;
  }
  for (int i = 0; i < q; i++) cout << ans[i] << "\n";
  return 0;
}
