#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
struct Ques {
  int l, r, id;
};
int n, bl, m, all, cnt;
int a[N];
Ques q[N];
deque<int> w[N];
deque<int> s[N];
int v[N], ans[N], ok[N];
int cmp(Ques a, Ques b) {
  return a.l / bl < b.l / bl || (a.l / bl == b.l / bl && a.r < b.r);
}
int main() {
  scanf("%d", &n);
  bl = sqrt(n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) scanf("%d%d", &q[i].l, &q[i].r);
  for (int i = 1; i <= m; i++) q[i].id = i;
  sort(q + 1, q + 1 + m, cmp);
  int l = 1, r = 0;
  for (int i = 1; i <= m; i++) {
    while (r < q[i].r) {
      r++, v[a[r]]++, cnt += v[a[r]] == 1;
      int p = a[r];
      all -= ok[p] == 0 && !s[p].empty();
      if (s[p].size() > 0) {
        if (!w[p].empty()) ok[p] += (r - s[p].back()) != w[p].back();
        w[p].push_back(r - s[p].back());
      }
      s[p].push_back(r);
      all += ok[p] == 0 && !s[p].empty();
    }
    while (q[i].l < l) {
      l--, v[a[l]]++, cnt += v[a[l]] == 1;
      int p = a[l];
      all -= ok[p] == 0 && !s[p].empty();
      if (s[p].size() > 0) {
        if (!w[p].empty()) ok[p] += (s[p].front() - l) != w[p].front();
        w[p].push_front(s[p].front() - l);
      }
      s[p].push_front(l);
      all += ok[p] == 0 && !s[p].empty();
    }
    while (l < q[i].l) {
      v[a[l]]--, cnt -= v[a[l]] == 0;
      int p = a[l];
      all -= ok[p] == 0 && !s[p].empty();
      s[p].pop_front();
      if (s[p].size() > 0) {
        int tmp = s[p].front() - l;
        w[p].pop_front();
        if (!w[p].empty()) ok[p] -= tmp != w[p].front();
      }
      all += ok[p] == 0 && !s[p].empty();
      l++;
    }
    while (q[i].r < r) {
      v[a[r]]--, cnt -= v[a[r]] == 0;
      int p = a[r];
      all -= ok[p] == 0 && !s[p].empty();
      s[p].pop_back();
      if (s[p].size() > 0) {
        int tmp = r - s[p].back();
        w[p].pop_back();
        if (!w[p].empty()) ok[p] -= tmp != w[p].back();
      }
      all += ok[p] == 0 && !s[p].empty();
      r--;
    }
    ans[q[i].id] = cnt + (all == 0);
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
struct Ques {
  int l, r, id;
};
int n, bl, m, all, cnt;
int a[N];
Ques q[N];
deque<int> w[N];
deque<int> s[N];
int v[N], ans[N], ok[N];
int cmp(Ques a, Ques b) {
  return a.l / bl < b.l / bl || (a.l / bl == b.l / bl && a.r < b.r);
}
int main() {
  scanf("%d", &n);
  bl = sqrt(n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) scanf("%d%d", &q[i].l, &q[i].r);
  for (int i = 1; i <= m; i++) q[i].id = i;
  sort(q + 1, q + 1 + m, cmp);
  int l = 1, r = 0;
  for (int i = 1; i <= m; i++) {
    while (r < q[i].r) {
      r++, v[a[r]]++, cnt += v[a[r]] == 1;
      int p = a[r];
      all -= ok[p] == 0 && !s[p].empty();
      if (s[p].size() > 0) {
        if (!w[p].empty()) ok[p] += (r - s[p].back()) != w[p].back();
        w[p].push_back(r - s[p].back());
      }
      s[p].push_back(r);
      all += ok[p] == 0 && !s[p].empty();
    }
    while (q[i].l < l) {
      l--, v[a[l]]++, cnt += v[a[l]] == 1;
      int p = a[l];
      all -= ok[p] == 0 && !s[p].empty();
      if (s[p].size() > 0) {
        if (!w[p].empty()) ok[p] += (s[p].front() - l) != w[p].front();
        w[p].push_front(s[p].front() - l);
      }
      s[p].push_front(l);
      all += ok[p] == 0 && !s[p].empty();
    }
    while (l < q[i].l) {
      v[a[l]]--, cnt -= v[a[l]] == 0;
      int p = a[l];
      all -= ok[p] == 0 && !s[p].empty();
      s[p].pop_front();
      if (s[p].size() > 0) {
        int tmp = s[p].front() - l;
        w[p].pop_front();
        if (!w[p].empty()) ok[p] -= tmp != w[p].front();
      }
      all += ok[p] == 0 && !s[p].empty();
      l++;
    }
    while (q[i].r < r) {
      v[a[r]]--, cnt -= v[a[r]] == 0;
      int p = a[r];
      all -= ok[p] == 0 && !s[p].empty();
      s[p].pop_back();
      if (s[p].size() > 0) {
        int tmp = r - s[p].back();
        w[p].pop_back();
        if (!w[p].empty()) ok[p] -= tmp != w[p].back();
      }
      all += ok[p] == 0 && !s[p].empty();
      r--;
    }
    ans[q[i].id] = cnt + (all == 0);
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}

```