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 long long INF = 1e9, MOD = INF + 7;
const int N = 1e5 + 5;
int n, a[N], fen[N], cnt[N], lstIdx[N], lst[N], dp[N], seg[4 * N];
vector<pair<int, int> > qR[N];
void add(int p, int x) {
  for (p++; p < N; p += p & -p) fen[p] += x;
}
int sum(int p) {
  int sum = 0;
  for (; p; p -= p & -p) sum += fen[p];
  return sum;
}
int sum(int l, int r) { return sum(r) - sum(l); }
void update(int p, int val, int id = 1, int s = 0, int e = n) {
  if (e - s < 2) {
    seg[id] = val;
    return;
  }
  int md = (s + e) / 2;
  if (p < md)
    update(p, val, 2 * id, s, md);
  else
    update(p, val, 2 * id + 1, md, e);
  seg[id] = min(seg[2 * id], seg[2 * id + 1]);
}
int get(int l, int r, int id = 1, int s = 0, int e = n) {
  if (l >= e || s >= r) return INF;
  if (l <= s && e <= r) return seg[id];
  int md = (s + e) / 2;
  return min(get(l, r, 2 * id, s, md), get(l, r, 2 * id + 1, md, e));
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  memset(lst, -1, sizeof lst);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    qR[r].push_back(make_pair(l, i));
  }
  for (int i = 0; i < n; i++) {
    lstIdx[i] = lst[a[i]];
    lst[a[i]] = i;
    if (lstIdx[i] == -1 || lstIdx[lstIdx[i]] == -1)
      dp[i] = -1;
    else if (i - lstIdx[i] == lstIdx[i] - lstIdx[lstIdx[i]])
      dp[i] = dp[lstIdx[i]];
    else
      dp[i] = lstIdx[lstIdx[i]];
  }
  for (int i = 0; i < n; i++) {
    if (lstIdx[i] != -1) add(lstIdx[i], -1), update(lstIdx[i], INF);
    add(i, +1);
    update(i, dp[i]);
    for (auto p : qR[i])
      cnt[p.second] = sum(p.first, i + 1) + (get(p.first, i + 1) >= p.first);
  }
  for (int i = 0; i < q; i++) cout << cnt[i] << "\n";
  return 0;
}
