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, int size>
struct fenwick {
  T a[size];
  void add(int pos, const T& val) {
    while (pos < size) {
      a[pos] += val;
      pos += pos & -pos;
    }
  }
  T sum(int pos) {
    T ret = T();
    while (pos > 0) {
      ret += a[pos];
      pos -= pos & -pos;
    }
    return ret;
  }
};
fenwick<int, 100005> drvo, marks;
int seen[100005];
int n;
int a[100005];
basic_string<int> e[100005];
int q, ql[100005], qr[100005], ans[100005];
basic_string<int> qe[100005];
struct job {
  int x, y;
};
basic_string<job> jobs[100005];
void add(int l1, int r1, int l2, int r2) {
  jobs[l2] += {l1, 1};
  jobs[l2] += {r1 + 1, -1};
  jobs[r2 + 1] += {l1, -1};
  jobs[r2 + 1] += {r1 + 1, 1};
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    e[a[i]] += i;
  }
  for (const auto& v : e) {
    if (!v.size()) continue;
    int j = 0;
    for (int i = 0; i < (int)v.size(); i++) {
      if (i && v[i] - v[i - 1] != v[j + 1] - v[j]) j = i - 1;
      int l2 = v[i];
      int r2 = i + 1 == (int)v.size() ? n : v[i + 1] - 1;
      int l1 = j == 0 ? 1 : v[j - 1] + 1;
      int r1 = v[i];
      add(l1, r1, l2, r2);
    }
  }
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> ql[i] >> qr[i];
    qe[qr[i]] += i;
  }
  for (int i = 1; i <= n; i++) {
    if (seen[a[i]]) drvo.add(seen[a[i]], -1);
    seen[a[i]] = i;
    drvo.add(i, 1);
    for (auto [x, y] : jobs[i]) marks.add(x, y);
    for (int j : qe[i]) {
      ans[j] = drvo.sum(i) - drvo.sum(ql[j] - 1);
      ans[j] += !marks.sum(ql[j]);
    }
  }
  for (int i = 0; i < q; i++) cout << ans[i] << '\n';
}
