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 N = 100000;
struct Binary_Indexed_Tree {
  int tree[N + 1];
  Binary_Indexed_Tree() {
    memset(tree, 0, sizeof(tree));
    return;
  }
  int lowbit(int x) { return x & -x; }
  void add(int x, int y) {
    if (x == 0) return;
    for (int i = x; i <= N; i += lowbit(i)) tree[i] += y;
    return;
  }
  void add(int a, int b, int val) {
    if (a > b) return;
    add(a, val);
    add(b + 1, -val);
    return;
  }
  int getsum(int x) {
    int res = 0;
    for (int i = x; i > 0; i -= lowbit(i)) res += tree[i];
    return res;
  }
} T1, T2;
struct Ask {
  int l, r, id;
  bool operator<(const Ask& b) const { return r < b.r; }
} query[N + 1];
int a[N + 1], book[N + 1];
int ans[N + 1];
vector<int> pos[N + 1];
int main() {
  int n, Q;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= N; i++) pos[i].push_back(0);
  scanf("%d", &Q);
  for (int i = 1; i <= Q; i++) {
    scanf("%d%d", &query[i].l, &query[i].r);
    query[i].id = i;
  }
  sort(query + 1, query + Q + 1);
  int j = 1;
  for (int i = 1; i <= n && j <= Q; i++) {
    int v = a[i];
    pos[v].push_back(i);
    int p = pos[v].size() - 1;
    T2.add(pos[v][p - 1] + 1, i, 1);
    if (p == 1)
      book[i] = 1;
    else {
      if (p == 2 || pos[v][p] - pos[v][p - 1] == pos[v][p - 1] - pos[v][p - 2])
        book[i] = book[pos[v][p - 1]];
      else {
        book[i] = pos[v][p - 2] + 1;
        T2.add(book[pos[v][p - 1]], book[i] - 1, -1);
      }
    }
    T1.add(pos[v][p - 1] + 1, i, 1);
    while (j <= Q && query[j].r == i) {
      ans[query[j].id] = T1.getsum(query[j].l);
      if (T2.getsum(query[j].l) == 0) ans[query[j].id]++;
      j++;
    }
  }
  for (int i = 1; i <= Q; i++) printf("%d\n", ans[i]);
  return 0;
}
