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 NMAX = 100000 + 5;
int N;
int b[NMAX];
int extend[NMAX];
vector<int> freq[NMAX];
int ans[NMAX];
vector<pair<int, int> > queries[NMAX];
void add(int aib[], int where, int val) {
  if (where == 0) return;
  for (; where <= N; where += where & (-where)) aib[where] += val;
}
int query(int aib[], int where) {
  int ans = 0;
  for (; where; where -= where & (-where)) ans += aib[where];
  return ans;
}
int query(int aib[], int l, int r) {
  int ans = query(aib, r);
  if (l > 0) ans -= query(aib, l - 1);
  return ans;
}
int aib_colors[NMAX];
int aib_constraints[NMAX];
int main() {
  ios_base ::sync_with_stdio(false);
  cin >> N;
  for (int i = 1; i <= N; ++i) cin >> b[i];
  int M;
  cin >> M;
  for (int i = 1; i <= M; ++i) {
    int l, r;
    cin >> l >> r;
    queries[r].push_back(make_pair(l, i));
  }
  for (int i = 1; i <= N; ++i) {
    const int val = b[i];
    vector<int> &v = freq[val];
    v.push_back(i);
    if (v.size() <= 2)
      extend[i] = 0;
    else {
      const int dif = v[v.size() - 1] - v[v.size() - 2];
      if (dif == v[v.size() - 2] - v[v.size() - 3])
        extend[i] = extend[v[v.size() - 2]];
      else
        extend[i] = v[v.size() - 3];
    }
    if (v.size() > 1) {
      add(aib_colors, v[v.size() - 2], -1);
      add(aib_constraints, extend[v[v.size() - 2]], -1);
    }
    add(aib_colors, i, 1);
    add(aib_constraints, extend[i], 1);
    for (auto it : queries[i]) {
      const int l = it.first;
      const int r = i;
      const int pos = it.second;
      int &anss = ans[pos];
      anss = query(aib_colors, l, r);
      if (query(aib_constraints, l, r) == anss) ++anss;
    }
  }
  for (int i = 1; i <= M; ++i) cout << ans[i] << '\n';
  return 0;
}
