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;
namespace FGF {
const int N = 1e5 + 5;
int n, m;
int a[N], bel[N], B;
struct ques {
  int l, r, i;
} q[N];
int cnt[N], sum, ans[N], fl, lst[N], c[N], len[N], tot, nxt[N];
bool cmp(const ques &a, const ques &b) {
  return bel[a.l] ^ bel[b.l] ? bel[a.l] < bel[b.l] : a.r < b.r;
}
deque<int> Q[N];
int askk(int x) {
  if (cnt[x] == 1) return 1;
  if (cnt[x] && c[nxt[Q[x].front()]] == c[Q[x].back()]) return 1;
  return 0;
}
void work() {
  scanf("%d", &n);
  B = sqrt(n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (!lst[a[i]])
      c[i] = ++tot;
    else
      (!len[a[i]] || len[a[i]] == i - lst[a[i]]) ? c[i] = c[lst[a[i]]]
                                                 : c[i] = ++tot;
    if (lst[a[i]]) nxt[lst[a[i]]] = i, len[a[i]] = i - lst[a[i]];
    lst[a[i]] = i, bel[i] = (i - 1) / B + 1;
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) scanf("%d%d", &q[i].l, &q[i].r), q[i].i = i;
  sort(q + 1, q + m + 1, cmp);
  for (int i = 1, l = 1, r = 0; i <= m; i++) {
    while (l > q[i].l)
      fl -= askk(a[--l]), sum += (++cnt[a[l]] == 1), Q[a[l]].push_front(l),
          fl += askk(a[l]);
    while (r < q[i].r)
      fl -= askk(a[++r]), sum += (++cnt[a[r]] == 1), Q[a[r]].push_back(r),
          fl += askk(a[r]);
    while (l < q[i].l)
      fl -= askk(a[l]), sum -= (--cnt[a[l]] == 0), Q[a[l]].pop_front(),
          fl += askk(a[l++]);
    while (r > q[i].r)
      fl -= askk(a[r]), sum -= (--cnt[a[r]] == 0), Q[a[r]].pop_back(),
          fl += askk(a[r--]);
    ans[q[i].i] = sum + (fl == 0);
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
}
}  // namespace FGF
int main() {
  FGF::work();
  return 0;
}
