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;
const int maxn = 1e5 + 10;
int n, m, num, oks, sum[maxn], sq, ans[maxn], l, r, a[maxn], f[maxn], cnt[maxn];
vector<int> v[maxn];
vector<pair<pair<int, int>, int> > que;
deque<int> dq[maxn];
bool cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  if (a.first.first / sq == b.first.first / sq) {
    if ((a.first.first / sq) % 2 == 0) return a.first.second < b.first.second;
    return a.first.second > b.first.second;
  }
  return a.first.first < b.first.first;
}
bool isok(int x, int y) { return ((y - x == 2 && x % 2 == 0) || y - x <= 1); }
bool isart(int a, int b, int c) { return 2 * b == a + c; }
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i], v[a[i]].push_back(i);
  sq = sqrt(n);
  for (int i = 0; i < maxn; i++) {
    int tmp = 0;
    for (int j = 1; j < v[i].size(); j++)
      if (j != v[i].size() - 1 && !isart(v[i][j - 1], v[i][j], v[i][j + 1]))
        f[v[i][j]] = tmp + 2, tmp += 2;
      else
        f[v[i][j]] = tmp + 1;
  }
  cin >> m;
  for (int i = 0, fi, se; i < m; i++)
    cin >> fi >> se, fi--, que.push_back({{fi, se}, i});
  sort(que.begin(), que.end(), cmp);
  for (auto q : que) {
    int s = q.first.first, e = q.first.second;
    while (s < l) {
      l--;
      int cur = a[l];
      if (!cnt[cur]++) num++;
      if (dq[cur].size()) oks -= isok(dq[cur].back(), dq[cur].front());
      dq[cur].push_back(f[l]), oks += isok(dq[cur].back(), dq[cur].front());
    }
    while (r < e) {
      int cur = a[r];
      if (!cnt[cur]++) num++;
      if (dq[cur].size()) oks -= isok(dq[cur].back(), dq[cur].front());
      dq[cur].push_front(f[r]), oks += isok(dq[cur].back(), dq[cur].front());
      r++;
    }
    while (l < s) {
      int cur = a[l];
      cnt[cur]--;
      if (!cnt[cur]) num--;
      oks -= isok(dq[cur].back(), dq[cur].front());
      dq[cur].pop_back();
      if (dq[cur].size()) oks += isok(dq[cur].back(), dq[cur].front());
      l++;
    }
    while (e < r) {
      r--;
      int cur = a[r];
      cnt[cur]--;
      if (!cnt[cur]) num--;
      oks -= isok(dq[cur].back(), dq[cur].front());
      dq[cur].pop_front();
      if (dq[cur].size()) oks += isok(dq[cur].back(), dq[cur].front());
    }
    if (oks)
      ans[q.second] = num;
    else
      ans[q.second] = num + 1;
  }
  for (int i = 0; i < m; i++) cout << ans[i] << "\n";
}
