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 Query {
  int l, r, id;
};
class MO {
  int cnt_arth;
  int distinct;
  vector<int> cnt, prg;
  vector<deque<int> > it;
  bool is_arth_prg(int num) const {
    if (it[num].empty()) return false;
    return (it[num].size() <= prg[it[num][0]]);
  }

 public:
  MO(vector<int> &ar) {
    cnt_arth = 0;
    distinct = 0;
    cnt.assign(100005, 0);
    prg.resize(100005);
    it.resize(100005);
    vector<vector<int> > idx(100005);
    for (int i = 0; i < ar.size(); i++) {
      idx[ar[i]].push_back(i);
    }
    for (int i = 0; i < 100005; i++) {
      if (idx[i].empty()) continue;
      prg[idx[i].back()] = 1;
      int d = -1;
      for (int j = idx[i].size() - 2; j >= 0; j--) {
        if (idx[i][j + 1] - idx[i][j] != d) {
          prg[idx[i][j]] = 2;
          d = idx[i][j + 1] - idx[i][j];
        } else
          prg[idx[i][j]] = prg[idx[i][j + 1]] + 1;
      }
    }
  }
  void add(int num, int idx, bool pole) {
    if (!cnt[num]) distinct++;
    cnt[num]++;
    bool bef = is_arth_prg(num);
    if (pole)
      it[num].push_front(idx);
    else
      it[num].push_back(idx);
    bool aft = is_arth_prg(num);
    if (!bef && aft)
      cnt_arth++;
    else if (bef && !aft)
      cnt_arth--;
  }
  void rmv(int num, bool pole) {
    cnt[num]--;
    if (!cnt[num]) distinct--;
    bool bef = is_arth_prg(num);
    if (pole)
      it[num].pop_front();
    else
      it[num].pop_back();
    bool aft = is_arth_prg(num);
    if (!bef && aft)
      cnt_arth++;
    else if (bef && !aft)
      cnt_arth--;
  }
  inline int answer() { return distinct + (cnt_arth ? 0 : 1); }
};
int main() {
  int n, q;
  scanf("%d", &n);
  vector<int> a(n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  MO mo(a);
  int block = sqrt(n + 1);
  scanf("%d", &q);
  vector<Query> qr(q);
  for (int i = 0; i < q; i++) {
    scanf("%d %d", &qr[i].l, &qr[i].r);
    qr[i].l--;
    qr[i].r--;
    qr[i].id = i;
  }
  sort((qr).begin(), (qr).end(),
       [block](const Query &a, const Query &b) -> bool {
         if (a.l / block == b.l / block) return (a.r < b.r);
         return (a.l / block < b.l / block);
       });
  vector<int> ans(q);
  int mo_left = 0, mo_right = -1;
  for (Query &x : qr) {
    while (mo_right < x.r) {
      mo.add(a[++mo_right], mo_right, 0);
    }
    while (mo_right > x.r) {
      mo.rmv(a[mo_right--], 0);
    }
    while (mo_left < x.l) {
      mo.rmv(a[mo_left++], 1);
    }
    while (mo_left > x.l) {
      mo.add(a[--mo_left], mo_left, 1);
    }
    ans[x.id] = mo.answer();
  }
  for (int i = 0; i < q; i++) printf("%d\n", ans[i]);
  return 0;
}
