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 = 100005;
int ft[N][2];
int get(int x, int id) {
  int ans = 0;
  for (; x > 0; x -= x & -x) ans += ft[x][id];
  return ans;
}
void upd(int x, int id, int val) {
  for (; x < N; x += x & -x) ft[x][id] += val;
}
void upd(int l, int r, int id, int val) {
  upd(l, id, val);
  upd(r + 1, id, -val);
}
int n, a[N];
int q;
vector<pair<int, int>> qu[N];
vector<int> occ[N];
int last[N], lastw[N];
int ans[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    qu[r].push_back({l, i});
  }
  for (int i = 1; i <= n; i++) {
    upd(last[a[i]] + 1, i, 0, 1);
    last[a[i]] = i;
    vector<int> &vec = occ[a[i]];
    if (vec.size() >= 2 and
        i - vec.back() != vec.back() - vec[vec.size() - 2]) {
      upd(lastw[a[i]] + 1, vec[vec.size() - 2], 1, 1);
      lastw[a[i]] = vec[vec.size() - 2];
    }
    vec.push_back(i);
    for (auto [l, id] : qu[i]) {
      ans[id] = get(l, 0);
      if (ans[id] == get(l, 1)) {
        ans[id]++;
      }
    }
  }
  for (int i = 1; i <= q; i++) {
    printf("%d\n", ans[i]);
  }
}
