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 n, a[111111], nex[111111], bod[111111], ind[111111], sz, m, x, y,
    ans[111111];
int bit1[222222], bit2[222222];
vector<pair<int, int> > que[111111];
void add(int p, int x, int *bit) {
  while (p <= sz) {
    *(bit + p) += x;
    p += p & (-p);
  }
}
int get(int p, int *bit) {
  int res = 0;
  while (p >= 1) {
    res += *(bit + p);
    p -= p & (-p);
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i <= 100000; i++) ind[i] = 1e9;
  for (int i = n; i >= 1; i--) {
    nex[i] = ind[a[i]];
    ind[a[i]] = i;
  }
  for (int i = n; i >= 1; i--) {
    if (nex[i] == 1e9)
      bod[i] = n + 1;
    else if (nex[i] - i == nex[nex[i]] - nex[i])
      bod[i] = bod[nex[i]];
    else {
      bod[i] = nex[nex[i]];
      if (bod[i] == 1e9) bod[i] = n + 1;
    }
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &x, &y);
    que[x].push_back(make_pair(y, i));
  }
  sz = 1;
  while (sz < n + 2) sz *= 2;
  for (int i = n; i >= 1; i--) {
    add(i, 1, bit1);
    if (nex[i] != 1e9) add(nex[i], -1, bit1);
    add(i, 1, bit2);
    add(bod[i], -1, bit2);
    if (nex[i] != 1e9) {
      add(nex[i], -1, bit2);
      add(bod[nex[i]], 1, bit2);
    }
    for (int j = 0; j < que[i].size(); j++) {
      int r = que[i][j].first, id = que[i][j].second;
      ans[id] = get(r, bit1);
      if (!get(r, bit2)) ans[id]++;
    }
  }
  for (int i = 1; i <= m; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
