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;
struct BIT {
  int n = 262192 - 1;
  int tree[262192];
  void setsize(int n) { this->n = n; }
  void set(int index, int val) {
    if (index <= 0) return;
    int curr = this->query(index, index);
    this->add(index, val - curr);
  }
  void add(int index, int val) {
    if (index <= 0) return;
    while (index <= n) {
      tree[index] += val;
      index = index + shift(index);
    }
  }
  int shift(int x) { return x & (-x); }
  int getsum(int index) {
    int ret = 0;
    while (index > 0) {
      ret += tree[index];
      index = index - shift(index);
    }
    return ret;
  }
  int query(int start, int end) {
    if (start > end) return 0;
    return getsum(end) - getsum(start - 1);
  }
};
struct qrs {
  int l, r, id;
};
qrs qs[262192];
BIT tree1, tree2;
int ans[262192];
int arr[262192];
int lst[262192];
vector<int> pos[262192];
int fail[262192];
static bool cmp(const qrs &a, const qrs &b) { return a.r < b.r; }
int main() {
  int i, m, q, j;
  scanf("%d", &m);
  tree1.setsize(m);
  tree2.setsize(m);
  for (i = 1; i <= m; i++) {
    scanf("%d", &arr[i]);
  }
  scanf("%d", &q);
  for (i = 1; i <= q; i++) {
    scanf("%d%d", &qs[i].l, &qs[i].r);
    qs[i].id = i;
  }
  sort(qs + 1, qs + 1 + q, cmp);
  for (i = 1, j = 1; i <= q; i++) {
    while (j <= m && j <= qs[i].r) {
      tree1.set(lst[arr[j]], 0);
      lst[arr[j]] = j;
      tree1.set(j, 1);
      if ((int)pos[arr[j]].size() == 1) {
        tree2.set(pos[arr[j]].back(), 0);
      } else if ((int)pos[arr[j]].size() > 1) {
        int cnt = (int)pos[arr[j]].size();
        tree2.set(pos[arr[j]][cnt - 1], 0);
        if (j - pos[arr[j]][cnt - 1] !=
            pos[arr[j]][cnt - 1] - pos[arr[j]][cnt - 2]) {
          tree2.set(fail[arr[j]], 0);
          tree2.set(pos[arr[j]][cnt - 2], -1);
          fail[arr[j]] = pos[arr[j]][cnt - 2];
        }
      }
      tree2.set(j, 1);
      pos[arr[j]].push_back(j);
      j++;
    }
    ans[qs[i].id] = tree1.query(qs[i].l, qs[i].r) +
                    (tree2.query(qs[i].l, qs[i].r) > 0 ? 0 : 1);
  }
  for (i = 1; i <= q; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
