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 MX = 100010;
int n, m, i, j, k, x, y, all, sbn, a[MX], p[MX], last[MX], cnt[MX], sb[MX],
    bad[MX], r[MX];
vector<int> b[MX], c[MX];
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  scanf("%d", &m);
  for (i = 0; i < m; i++) {
    scanf("%d%d", &x, &y);
    b[y].push_back(x);
    c[y].push_back(i);
  }
  for (i = 1; i <= n; i++) {
    if (last[a[i]] > 0) {
      for (k = last[a[i]]; k < MX; k = (k << 1) - (k & (k - 1))) cnt[k]--;
      if (i - last[a[i]] != last[a[i]] - p[last[a[i]]]) {
        if (sb[a[i]] > 0) {
          for (k = sb[a[i]]; k < MX; k = (k << 1) - (k & (k - 1))) bad[k]--;
        } else if (p[last[a[i]]] > 0)
          sbn++;
        sb[a[i]] = p[last[a[i]]];
        if (sb[a[i]] > 0)
          for (k = sb[a[i]]; k < MX; k = (k << 1) - (k & (k - 1))) bad[k]++;
      }
    } else
      all++;
    p[i] = last[a[i]];
    last[a[i]] = i;
    for (k = i; k < MX; k = (k << 1) - (k & (k - 1))) cnt[k]++;
    for (j = 0; j < b[i].size(); j++) {
      int diff = all, scur = sbn;
      for (k = b[i][j] - 1; k > 0; k &= k - 1) {
        diff -= cnt[k];
        scur -= bad[k];
      }
      r[c[i][j]] = diff + int(scur == diff);
    }
  }
  for (i = 0; i < m; i++) printf("%d\n", r[i]);
  return 0;
}
