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>
#pragma warning(disable : 4996)
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 5;
int m, q;
int val[maxn], sum1[maxn], sum2[maxn], las[maxn], pre[maxn], fl[maxn],
    res[maxn];
struct no {
  int le, ri, id;
} qu[maxn];
bool cmp(no n1, no n2) { return n1.ri < n2.ri; }
void solve() {
  int i;
  scanf("%d", &m);
  for (i = 1; i <= m; i++) {
    scanf("%d", &val[i]);
    pre[i] = las[val[i]];
    las[val[i]] = i;
    if (i - pre[i] == pre[i] - pre[pre[i]]) {
      fl[i] = fl[pre[i]];
    } else {
      fl[i] = pre[pre[i]];
    }
  }
  scanf("%d", &q);
  for (i = 1; i <= q; i++) {
    scanf("%d%d", &qu[i].le, &qu[i].ri);
    qu[i].id = i;
  }
  sort(qu + 1, qu + q + 1, cmp);
  int j, k = 1;
  for (i = 1; i <= m; i++) {
    if (i == 10) {
      i++;
      i--;
    }
    for (j = i; j; j -= (j & (-j))) {
      sum1[j]++;
    }
    for (j = pre[i]; j; j -= (j & (-j))) {
      sum1[j]--;
    }
    for (j = fl[i]; j; j -= (j & (-j))) {
      sum2[j]++;
    }
    for (j = fl[pre[i]]; j; j -= (j & (-j))) {
      sum2[j]--;
    }
    while (qu[k].ri == i && k <= q) {
      int su1 = 0, su2 = 0;
      for (j = qu[k].le; j <= i; j += (j & (-j))) {
        su1 += sum1[j];
        su2 += sum2[j];
      }
      res[qu[k].id] = su1 + (su1 == su2);
      k++;
    }
  }
  for (i = 1; i <= q; i++) {
    printf("%d\n", res[i]);
  }
}
int main() {
  solve();
  return 0;
}
