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>
const int N = 100005;
int n, m;
int c[2][N], ans[N];
std::vector<int> v[N], d[N];
int a[N], lef[N], las[N], P[N];
void Insert(int t, int x, int d) {
  for (; x; x -= x & -x) c[t][x] += d;
}
int Query(int t, int x) {
  int ret = 0;
  for (; x <= n; x += x & -x) ret += c[t][x];
  return ret;
}
int main() {
  int i, j, x, y;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    lef[i] = las[a[i]];
    las[a[i]] = i;
    if (i - lef[i] == lef[i] - lef[lef[i]])
      P[i] = P[lef[i]];
    else
      P[i] = lef[lef[i]];
  }
  scanf("%d", &m);
  for (i = 1; i <= m; i++)
    scanf("%d%d", &x, &y), v[y].push_back(x), d[y].push_back(i);
  for (i = 1; i <= n; i++) {
    if (lef[i])
      Insert(0, lef[i], -1), Insert(1, P[lef[i]], 1), Insert(1, lef[i], -1);
    Insert(0, i, 1);
    Insert(1, P[i], -1);
    Insert(1, i, 1);
    for (j = 0; j < v[i].size(); j++)
      ans[d[i][j]] = Query(0, v[i][j]) + !Query(1, v[i][j]);
  }
  for (i = 1; i <= m; i++) printf("%d\n", ans[i]);
}
