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 = 1e5 + 10;
int id[N], pre[N], aa[N], ll[N];
int cc[2][N], dd[N];
int n;
void update(int x, int v, int id = 0) {
  if (x == 0) return;
  while (x <= n) cc[id][x] += v, x += x & -x;
}
int query(int x, int id = 0) {
  int ans = 0;
  while (x) ans += cc[id][x], x -= x & -x;
  return ans;
}
struct PP {
  int l, r, id;
  bool operator<(const PP& th) const { return r < th.r; }
} pp[N];
int ans[N];
int main() {
  while (scanf("%d", &n) != EOF) {
    for (int i = 1; i <= n; i++) scanf("%d", &aa[i]);
    int q;
    scanf("%d", &q);
    for (int i = 0; i < q; i++) scanf("%d%d", &pp[i].l, &pp[i].r), pp[i].id = i;
    sort(pp, pp + q);
    memset(id, 0, sizeof(id));
    memset(pre, 0, sizeof(pre));
    memset(cc, 0, sizeof(cc));
    for (int i = 1, j = 0; i <= n; i++) {
      pre[i] = id[aa[i]];
      id[aa[i]] = i;
      update(pre[i], -1);
      update(i, 1);
      if (pre[pre[i]] == 0 || i - pre[i] == pre[i] - pre[pre[i]])
        ll[i] = ll[pre[i]];
      else
        ll[i] = pre[i];
      update(pre[ll[i]], -1, 1);
      update(i, 1, 1);
      update(pre[ll[pre[i]]], 1, 1);
      update(pre[i], -1, 1);
      while (j < q && pp[j].r == i) {
        ans[pp[j].id] = query(pp[j].r) - query(pp[j].l - 1) +
                        ((query(pp[j].r, 1) - query(pp[j].l - 1, 1)) == 0);
        j++;
      }
    }
    for (int i = 0; i < q; i++) printf("%d\n", ans[i]);
  }
  return 0;
}
