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 N = 1000 * 100 + 100, INF = 1000 * 1000 * 1000;
int dp[N], b[N], ans[N], seg[4 * N], n, t[4 * N];
vector<int> num[N];
vector<pair<int, int> > q[N];
void upd(int p, int val, int s = 0, int e = n, int id = 1) {
  if (e - s == 1) {
    seg[id] = val;
    return;
  }
  int mid = (s + e) / 2;
  if (p < mid)
    upd(p, val, s, mid, id * 2);
  else
    upd(p, val, mid, e, id * 2 + 1);
  seg[id] = min(seg[id * 2], seg[id * 2 + 1]);
}
int get(int l, int r, int s = 0, int e = n, int id = 1) {
  if (r <= s || e <= l) return INF;
  if (l <= s && e <= r) return seg[id];
  int mid = (s + e) / 2;
  return min(get(l, r, s, mid, id * 2), get(l, r, mid, e, id * 2 + 1));
}
void updt(int p, int val, int s = 0, int e = n, int id = 1) {
  t[id] += val;
  if (e - s == 1) return;
  int mid = (s + e) / 2;
  if (p < mid)
    updt(p, val, s, mid, id * 2);
  else
    updt(p, val, mid, e, id * 2 + 1);
}
int get2(int l, int r, int s = 0, int e = n, int id = 1) {
  if (r <= s || e <= l) return 0;
  if (l <= s && e <= r) return t[id];
  int mid = (s + e) / 2;
  return get2(l, r, s, mid, id * 2) + get2(l, r, mid, e, id * 2 + 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> b[i];
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    q[r].push_back({l, i});
  }
  for (int i = 0; i < N; i++) dp[i] = INF;
  for (int i = 0; i < 4 * N; i++) seg[i] = INF;
  for (int i = 0; i < n; i++) {
    num[b[i]].push_back(i);
    int s = num[b[i]].size();
    if (s == 1)
      dp[i] = 0;
    else if (s == 2) {
      dp[i] = 0;
      dp[num[b[i]][s - 2]] = INF;
      upd(num[b[i]][s - 2], INF);
      updt(num[b[i]][s - 2], -1);
    } else {
      if (num[b[i]][s - 1] - num[b[i]][s - 2] ==
          num[b[i]][s - 2] - num[b[i]][s - 3])
        dp[i] = dp[num[b[i]][s - 2]];
      else
        dp[i] = num[b[i]][s - 3] + 1;
      dp[num[b[i]][s - 2]] = INF;
      upd(num[b[i]][s - 2], INF);
      updt(num[b[i]][s - 2], -1);
    }
    upd(i, dp[i]);
    updt(i, +1);
    for (int j = 0; j < q[i].size(); j++) {
      int l = q[i][j].first;
      ans[q[i][j].second] = get2(l, i + 1);
      if (get(l, i + 1) > l) ans[q[i][j].second]++;
    }
  }
  for (int i = 0; i < m; i++) cout << ans[i] << '\n';
  return 0;
}
