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;
using PII = pair<int, int>;
using TPL = pair<PII, int>;
class BITree {
 public:
  static const int SIZE = 100010, BIAS = 5;
  int u[SIZE];
  void modify(int x, int v) {
    for (x += BIAS; x < SIZE; x += x & -x) u[x] += v;
  }
  int getsum(int x) {
    int s = 0;
    for (x += BIAS; x; x -= x & -x) s += u[x];
    return s;
  }
} s, g;
int o[100005];
int v[100005], gap[100005];
int a[100005], ans[100005];
TPL u[100005];
int main() {
  int n, m;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    u[i] = {{y - 1, x - 1}, i};
  }
  sort(u, u + m);
  u[m] = {{n, n}, m};
  memset(v, -1, sizeof(v));
  memset(o, -1, sizeof(o));
  for (int i = 0, j = 0; i < n; i++) {
    if (~v[a[i]]) {
      s.modify(v[a[i]], -1);
      gap[i] = i - v[a[i]];
      if (gap[i] != gap[v[a[i]]] && gap[v[a[i]]]) {
        if (~o[a[i]]) g.modify(o[a[i]], -1);
        g.modify(o[a[i]] = v[a[i]] - gap[v[a[i]]], 1);
      }
    }
    s.modify(v[a[i]] = i, 1);
    while (i == u[j].first.first) {
      int cnt = s.getsum(i) - s.getsum(u[j].first.second - 1);
      int tag = g.getsum(i) - g.getsum(u[j].first.second - 1);
      ans[u[j++].second] = cnt + (tag == cnt);
    }
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
}
