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 maxn = 100003;
int n, a[maxn], lst[maxn], fa[maxn], ffa[maxn], T_T, now, ans[maxn];
struct query {
  int l, r, id;
} q[maxn];
bool cmp(query x, query y) { return x.r < y.r; }
struct bit {
  int sum[maxn];
  void update(int x, int w) {
    if (x == 0) return;
    for (; x < maxn; x += ((x) & (-x))) {
      sum[x] += w;
    }
  }
  int query(int l, int r) {
    int res1 = 0, res2 = 0;
    --l;
    for (; l > 0; l -= ((l) & (-l))) res1 += sum[l];
    for (; r > 0; r -= ((r) & (-r))) res2 += sum[r];
    return res2 - res1;
  }
} cnt, ch;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    fa[i] = lst[a[i]];
    lst[a[i]] = i;
    if (fa[i] - fa[fa[i]] == i - fa[i])
      ffa[i] = ffa[fa[i]];
    else
      ffa[i] = fa[fa[i]];
  }
  scanf("%d", &T_T);
  for (int i = 1; i <= T_T; ++i) scanf("%d%d", &q[i].l, &q[i].r), q[i].id = i;
  sort(q + 1, q + T_T + 1, cmp);
  for (int i = 1; i <= T_T; ++i) {
    int l = q[i].l, r = q[i].r;
    while (now < r) {
      ++now;
      cnt.update(fa[now], -1);
      cnt.update(now, 1);
      ch.update(ffa[fa[now]], -1);
      ch.update(ffa[now], 1);
    }
    int t = cnt.query(l, r), tt = ch.query(l, r);
    ans[q[i].id] = t + (t == tt);
  }
  for (int i = 1; i <= T_T; ++i) printf("%d\n", ans[i]);
  return 0;
}
