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;
deque<int> d[100005];
int v[100005];
int p[100005];
pair<int, int> q[100005];
int ans = 0;
int prog = 0;
int as[100005];
unordered_map<int, int> possible[100005];
bool cmp(int a, int b) {
  if (q[a].first / 300 != q[b].first / 300)
    return q[a].first / 300 < q[b].first / 300;
  return q[a].second < q[b].second;
}
void update(int val, int pos, int u, int c) {
  if (c == -1) {
    prog -= possible[val].size() <= 1;
    if (u == 1) {
      int cnt = d[val].front();
      d[val].pop_front();
      if (d[val].empty())
        --ans;
      else {
        --possible[val][d[val].front() - cnt];
        if (!possible[val][d[val].front() - cnt])
          possible[val].erase(d[val].front() - cnt);
      }
    } else {
      int cnt = d[val].back();
      d[val].pop_back();
      if (d[val].empty())
        --ans;
      else {
        --possible[val][cnt - d[val].back()];
        if (!possible[val][cnt - d[val].back()])
          possible[val].erase(cnt - d[val].back());
      }
    }
    if (!d[val].empty()) prog += possible[val].size() <= 1;
  } else {
    if (d[val].empty())
      ++ans;
    else
      prog -= possible[val].size() <= 1;
    if (u == 1) {
      if (d[val].empty()) {
        d[val].push_front(pos);
      } else {
        int cnt = d[val].front();
        d[val].push_front(pos);
        ++possible[val][cnt - pos];
      }
    } else {
      if (d[val].empty()) {
        d[val].push_back(pos);
      } else {
        int cnt = d[val].back();
        d[val].push_back(pos);
        ++possible[val][pos - cnt];
      }
    }
    if (possible[val].size() <= 1) ++prog;
  }
}
int main(void) {
  int n, m;
  ios_base ::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> v[i];
  cin >> m;
  for (int i = 1; i <= m; ++i) cin >> q[i].first >> q[i].second, p[i] = i;
  sort(p + 1, p + 1 + m, cmp);
  for (int i = q[p[1]].first; i <= q[p[1]].second; ++i) update(v[i], i, 1, 1);
  as[p[1]] = ans + !prog;
  int l = q[p[1]].first, r = q[p[1]].second;
  for (int i = 2; i <= m; ++i) {
    while (l > q[p[i]].first) --l, update(v[l], l, -1, 1);
    while (r < q[p[i]].second) ++r, update(v[r], r, 1, 1);
    while (l < q[p[i]].first) update(v[l], l, -1, -1), ++l;
    while (r > q[p[i]].second) update(v[r], r, 1, -1), --r;
    as[p[i]] = ans + !prog;
  }
  for (int i = 1; i <= m; ++i) cout << as[i] << '\n';
  return cout << '\n', 0;
}
