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 <typename Tp>
inline void outarr(Tp _begin, Tp _end, const char* _delim = " ") {
  for (Tp current = _begin; current != _end; ++current) {
    std::cout << *current << _delim;
  }
  std::cout << '\n';
}
using ll = long long;
using pii = std::pair<int, int>;
constexpr int INF = 0x3f3f3f3f;
constexpr int MOD = static_cast<const int>(1e9 + 7);
constexpr int MAXN = 100000;
struct Query {
  int L;
  int R;
  int ID;
};
bool operator<(const Query& lhs, const Query& rhs) { return lhs.R < rhs.R; }
class FenwickTree {
 public:
  explicit FenwickTree(int n) : tree(n, 0) {}
  void Update(int i, int x) {
    for (; i < static_cast<int>((tree).size()); i |= i + 1) {
      tree[i] += x;
    }
  }
  int Sum(int r) {
    int res = 0;
    for (; r >= 0; r = (r & (r + 1)) - 1) {
      res += tree[r];
    }
    return res;
  }
  int Sum(int l, int r) { return Sum(r) - Sum(l - 1); }

 private:
  vector<int> tree;
};
class RangeFenwickTree {
 public:
  explicit RangeFenwickTree(int n) : ft(n) {}
  void Update(int l, int r, int x) {
    ft.Update(l, x);
    ft.Update(r + 1, -x);
  }
  int Get(int i) { return ft.Sum(i); }

 private:
  FenwickTree ft;
};
int arr[MAXN];
int ans[MAXN];
int delta[MAXN];
int last[MAXN];
int start[MAXN];
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 0; i < (n); ++i) {
    cin >> arr[i];
    --arr[i];
  }
  int q;
  cin >> q;
  vector<Query> query(q);
  for (int i = 0; i < (q); ++i) {
    cin >> query[i].L >> query[i].R;
    --query[i].L;
    --query[i].R;
    query[i].ID = i;
  }
  sort((query).begin(), (query).end());
  FenwickTree ft(n);
  RangeFenwickTree rft(n);
  memset(last, -1, sizeof(last));
  int cur = -1;
  for (auto [l, r, i] : query) {
    while (cur < r) {
      ++cur;
      if (last[arr[cur]] != -1) {
        if (cur - last[arr[cur]] != delta[arr[cur]]) {
          if (delta[arr[cur]] > 0) {
            rft.Update(start[arr[cur]], last[arr[cur]] - delta[arr[cur]], -1);
            start[arr[cur]] = last[arr[cur]] - delta[arr[cur]] + 1;
          }
          delta[arr[cur]] = cur - last[arr[cur]];
        }
        ft.Update(last[arr[cur]], -1);
      }
      rft.Update(last[arr[cur]] + 1, cur, 1);
      ft.Update(cur, 1);
      last[arr[cur]] = cur;
    }
    ans[i] = ft.Sum(l, r) + (rft.Get(l) == 0);
  }
  outarr(ans, ans + q, "\n");
  return 0;
}
