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 MAXM = 100 * 1000 + 123;
int m, b[MAXM], prv[MAXM], lst[MAXM], lbd[MAXM], l[MAXM], ans[MAXM],
    fen[2 + (1)][MAXM], q;
vector<int> v[MAXM];
void inp();
void pre();
void solve();
int fget(int, int);
void fadd(int, int, int);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  inp();
  pre();
  solve();
  for (int i = 0; i < q; i++) cout << ans[i] << '\n';
  return 0;
}
void inp() {
  cin >> m;
  for (int i = 0; i < m; i++) cin >> b[i];
  cin >> q;
  for (int i = 0; i < q; i++) {
    int r;
    cin >> l[i] >> r;
    v[r].push_back(i);
    l[i]--;
  }
}
void pre() {
  memset(lst, -1, sizeof lst);
  memset(lbd, -1, sizeof lbd);
  for (int i = 0; i < m; i++) {
    prv[i] = lst[b[i]];
    lst[b[i]] = i;
  }
  for (int i = 0; i < m; i++) {
    if (~prv[i]) lbd[i] = lbd[prv[i]];
    if ((~prv[i]) && (~prv[prv[i]]) && (i - prv[i] != prv[i] - prv[prv[i]]))
      lbd[i] = prv[prv[i]];
  }
}
void solve() {
  int tmp;
  for (int i = 1; i <= m; i++) {
    if (~prv[i - 1]) fadd(prv[i - 1], 0, -1);
    fadd(i - 1, 0, 1);
    if (~prv[i - 1] && ~lbd[prv[i - 1]]) fadd(lbd[prv[i - 1]], 1, -1);
    if (~lbd[i - 1]) fadd(lbd[i - 1], 1, 1);
    for (auto j : v[i]) {
      tmp = fget(i, 0) - fget(l[j], 0);
      ans[j] = tmp;
      ans[j] += (((fget(i, 1) - fget(l[j], 1)) < tmp) ? 0 : 1);
    }
  }
}
int fget(int p, int w) {
  int ret = 0;
  for (; p; p -= (p & -p)) ret += fen[w][p];
  return ret;
}
void fadd(int p, int w, int val) {
  for (p++; p <= m; p += (p & -p)) fen[w][p] += val;
}
