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;
template <class T>
inline void smin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void smax(T &a, T b) {
  if (a < b) a = b;
}
const int maxn = 100000 + 100;
int n, m;
int a[maxn];
vector<int> c[maxn];
vector<pair<int, int> > q[maxn];
int id[maxn], pid[maxn];
int f1[maxn], f2[maxn];
int ans[maxn];
void add(int *f, int x, int v) {
  for (x++; x < maxn; x += x & -x) f[x] += v;
}
int get(int *f, int x) {
  int r = 0;
  for (; x; x -= x & -x) r += f[x];
  return r;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0, _n = (int)(n); i < _n; i++) cin >> a[i], c[a[i]].push_back(i);
  cin >> m;
  for (int i = 0, _n = (int)(m); i < _n; i++) {
    int l, r;
    cin >> l >> r;
    q[--l].push_back({r, i});
  }
  for (int i = 0, _n = (int)(maxn); i < _n; i++)
    if ((int((c[i]).size()))) {
      vector<int> &C = c[i];
      int &P = pid[i];
      add(f1, C[0], 1);
      P = 2;
      while (P < (int((C).size())) && C[P] - C[P - 1] == C[P - 1] - C[P - 2])
        P++;
      if (P < (int((C).size()))) add(f2, C[P], 1);
    }
  for (int i = 0, _n = (int)(n); i < _n; i++) {
    for (auto qry : q[i])
      ans[qry.second] = get(f1, qry.first) +
                        (get(f2, qry.first) < get(f1, qry.first) ? 0 : 1);
    add(f1, i, -1);
    vector<int> &C = c[a[i]];
    int &I = id[a[i]], &P = pid[a[i]];
    I++;
    if (I < (int((C).size()))) add(f1, C[I], 1);
    if (C[P - 2] == i && P < (int((C).size()))) {
      add(f2, C[P], -1);
      P++;
      while (P < (int((C).size())) && C[P] - C[P - 1] == C[P - 1] - C[P - 2])
        P++;
      if (P < (int((C).size()))) add(f2, C[P], 1);
    }
  }
  for (int i = 0, _n = (int)(m); i < _n; i++) cout << ans[i] << endl;
  { return 0; }
}
