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;
bool cmp(const pair<pair<int, int>, int>& a,
         const pair<pair<int, int>, int>& b) {
  return a.first.second < b.first.second;
}
int cnt1[N], cnt2[N];
void add(int x, int val, int c[]) {
  for (; x < N; x += x & -x) {
    c[x] += val;
  }
}
int sum(int x, int c[]) {
  int res;
  for (res = 0; x; x -= x & -x) {
    res += c[x];
  }
  return res;
}
int a[N];
pair<pair<int, int>, int> Q[N];
pair<int, int> ans[N];
int pre[N], delta[N], put[N];
int main() {
  int n, i, m, j;
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) scanf("%d", &a[i]);
  scanf("%d", &m);
  for (i = 1; i <= m; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    Q[i] = make_pair(make_pair(l, r), i);
  }
  sort(Q + 1, Q + m + 1, cmp);
  memset(delta, -1, sizeof(delta));
  memset(pre, 0, sizeof(pre));
  int last = 0, mark;
  for (i = 1; i <= m; i = j) {
    mark = Q[i].first.second;
    for (j = last + 1; j <= mark; ++j) {
      if (!pre[a[j]]) {
        add(1, 1, cnt1);
        add(1, 1, cnt2);
        add(j + 1, -1, cnt1);
        add(j + 1, -1, cnt2);
        pre[a[j]] = j;
        put[a[j]] = 1;
      } else {
        add(pre[a[j]] + 1, 1, cnt1);
        add(pre[a[j]] + 1, 1, cnt2);
        add(j + 1, -1, cnt1);
        add(j + 1, -1, cnt2);
        if (delta[a[j]] == -1 || delta[a[j]] == j - pre[a[j]]) {
          delta[a[j]] = j - pre[a[j]];
        } else {
          add(put[a[j]], -1, cnt2);
          add(put[a[j]] = pre[a[j]] - delta[a[j]] + 1, 1, cnt2);
          delta[a[j]] = j - pre[a[j]];
        }
        pre[a[j]] = j;
      }
    }
    for (j = i; j <= m && Q[j].first.second == mark; ++j) {
      int tmp1, tmp2;
      int res = 1 + (tmp1 = sum(Q[j].first.first, cnt1)) -
                ((tmp2 = sum(Q[j].first.first, cnt2)) >= 1);
      ans[j] = make_pair(Q[j].second, res);
    }
    last = mark;
  }
  sort(ans + 1, ans + m + 1);
  for (i = 1; i <= m; ++i) printf("%d\n", ans[i].second);
  return 0;
}
