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 n, q;
int query(int *a, int i) {
  if (i < 0) return 0;
  int res = 0;
  while (i) {
    res += a[i];
    i -= (i & (-i));
  }
  return res;
}
int query(int *a, int l, int r) { return query(a, r) - query(a, l - 1); }
void add(int *a, int i, int del) {
  if (i <= 0) return;
  while (i <= n) {
    a[i] += del;
    i += (i & (-i));
  }
}
void add(int *a, int l, int r, int del) {
  add(a, l, del);
  add(a, r + 1, -del);
}
vector<pair<int, int> > qry[100010];
int a[100010];
int lst[100010], lst2[100010];
int tmp[100010];
int fir[100010];
int per[100010];
int can[100010], cnt[100010];
int res[100010];
int main(void) {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    lst[i] = tmp[a[i]];
    lst2[i] = lst[lst[i]];
    tmp[a[i]] = i;
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    qry[r].push_back(make_pair(l, i));
  }
  for (int i = 1; i <= n; ++i) {
    if (i - lst[i] != per[a[i]]) {
      if (!lst[i]) {
        add(can, 1, i, 1);
      } else {
        if (!lst2[i]) {
          per[a[i]] = i - lst[i];
          add(can, lst[i] + 1, i, 1);
        } else {
          add(can, fir[a[i]] + 1, lst[i], -1);
          fir[a[i]] = lst2[i];
          per[a[i]] = i - lst[i];
          add(can, fir[a[i]] + 1, i, 1);
        }
      }
    } else {
      add(can, lst[i] + 1, i, 1);
    }
    add(cnt, lst[i] + 1, 1);
    add(cnt, i + 1, -1);
    for (int j = qry[i].size() - 1; ~j; --j) {
      res[qry[i][j].second] = query(cnt, qry[i][j].first) + 1 -
                              ((query(can, qry[i][j].first) > 0) ? 1 : 0);
    }
  }
  for (int i = 1; i <= q; ++i) {
    printf("%d\n", res[i]);
  }
  return 0;
}
