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;
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1}, dy[] = {-1, 1, 0, 0, 1, -1, -1, 1};
void run();
int main() {
  ios::sync_with_stdio(0);
  cout << fixed << setprecision(16);
  run();
}
const int maxn = 100100;
int n, q;
int seq[maxn];
tuple<int, int, int> queries[maxn];
int last[maxn], a[maxn], asdasdasdasd[maxn], res[maxn];
int t1[maxn], t2[maxn];
int read_prefix(int tree[], int idx) {
  int sum = 0;
  for (; idx > 0; idx -= idx & -idx) sum += tree[idx];
  return sum;
}
void update_idx(int tree[], int idx, int val) {
  for (; idx <= n; idx += idx & -idx) tree[idx] += val;
}
int read_idx(int tree[], int idx) {
  int sum = 0;
  for (; idx <= n; idx += idx & -idx) sum += tree[idx];
  return sum;
}
void update_prefix(int tree[], int idx, int val) {
  for (; idx > 0; idx -= idx & -idx) tree[idx] += val;
}
int query_interv(int x) { return read_idx(t2, x); }
int query_last(int x) { return read_prefix(t1, x); }
void remove(int a, int last) {
  update_idx(t1, last, -1);
  update_prefix(t2, last, -1);
  update_prefix(t2, a - 1, +1);
}
void insert(int a, int last) {
  update_idx(t1, last, +1);
  update_prefix(t2, last, +1);
  update_prefix(t2, a - 1, -1);
}
void run() {
  cin >> n;
  for (int i = (1); i < (n + 1); ++i) cin >> seq[i];
  cin >> q;
  for (int i = (0); i < (q); ++i) {
    cin >> get<1>(queries[i]) >> get<0>(queries[i]);
    get<2>(queries[i]) = i;
  }
  sort(queries, queries + q);
  memset(last, -1, sizeof last);
  memset(asdasdasdasd, -1, sizeof asdasdasdasd);
  memset(a, -1, sizeof a);
  int i = 0;
  for (int qq = (0); qq < (q); ++qq) {
    int r = get<0>(queries[qq]), l = get<1>(queries[qq]);
    while (i < r) {
      int x = seq[++i];
      int j = last[x];
      if (j != -1) remove(a[x], j);
      if (j == -1 || asdasdasdasd[j] == -1) {
        a[x] = 1;
      } else if (i - j != j - asdasdasdasd[j]) {
        a[x] = asdasdasdasd[j] + 1;
      }
      asdasdasdasd[i] = j;
      last[x] = i;
      insert(a[x], i);
    }
    int ans = 1 + query_last(n) - query_last(l - 1);
    if (query_interv(l)) ans--;
    res[get<2>(queries[qq])] = ans;
  }
  for (int qq = (0); qq < (q); ++qq) cout << res[qq] << endl;
}
