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 = 100 * 1000 + 3333;
int good[N], a[N], bad[N], nxt[N], l[N], answer[N];
int r[N], inx[N], SegD[4 * N], SegT[4 * N];
void add(int *seg, int p, int val) {
  for (p++; p < N; p += p & -p) seg[p] += val;
}
int get(int *seg, int p) {
  int ans = 0;
  for (; p; p -= p & -p) ans += seg[p];
  return ans;
}
int get(int *seg, int l, int r) { return get(seg, r) - get(seg, l); }
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  int q;
  cin >> q;
  vector<int> vec[N];
  for (int i = 0; i < q; i++) {
    cin >> l[i] >> r[i];
    l[i]--;
    vec[l[i]].push_back(i);
  }
  for (int i = 0; i < N; i++) nxt[i] = n;
  for (int i = n - 1; i >= 0; i--) {
    bad[i] = nxt[a[i]];
    if (bad[i] != n) {
      add(SegD, bad[i], -1);
      add(SegT, bad[i], -1);
      add(SegD, good[bad[i]], 1);
    }
    add(SegD, i, 1);
    add(SegT, i, 1);
    if (bad[i] == n || bad[bad[i]] == n) {
      good[i] = n;
    } else {
      if (bad[i] - i == bad[bad[i]] - bad[i])
        good[i] = good[bad[i]];
      else
        good[i] = bad[bad[i]];
    }
    nxt[a[i]] = i;
    add(SegD, good[i], -1);
    for (int v : vec[i]) {
      int ans = get(SegT, l[v], r[v]);
      if (!get(SegD, l[v], r[v])) ans++;
      answer[v] = ans;
    }
  }
  for (int i = 0; i < q; i++) cout << answer[i] << endl;
}
