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 a[100005];
struct dat {
  int l, r, t;
  dat(int l = 0, int r = 0, int t = 0) : l(l), r(r), t(t) {}
};
dat q[100005];
int b[100005], p[100005];
bool operator<(dat a, dat b) { return a.r < b.r; }
void add(int *a, int b, int c) {
  for (; b < 100005; b += b & -b) a[b] += c;
}
int sum(int *a, int b) {
  int res = 0;
  for (; b; b -= b & -b) res += a[b];
  return res;
}
int A[100005], B[100005], res[100005];
int main() {
  int n, i;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    p[i] = b[a[i]];
    b[a[i]] = i;
  }
  int m;
  scanf("%d", &m);
  int st, en;
  for (i = 1; i <= m; i++) {
    scanf("%d %d", &st, &en);
    q[i] = dat(st, en, i);
  }
  sort(q + 1, q + m + 1);
  int j = 1;
  memset(b, 0, sizeof(b));
  for (i = 1; i <= n; i++) {
    add(A, p[i] + 1, 1);
    add(A, i + 1, -1);
    if (p[p[i]] && i - p[i] != p[i] - p[p[i]]) {
      add(B, b[a[i]] + 1, 1);
      add(B, p[p[i]] + 1, -1);
      b[a[i]] = p[p[i]];
    }
    while (j <= m && q[j].r < i) j++;
    while (j <= m && q[j].r == i) {
      int s1 = sum(A, q[j].l);
      int s2 = sum(B, q[j].l);
      if (s2 == s1) s1++;
      res[q[j].t] = s1;
      j++;
    }
  }
  for (i = 1; i <= m; i++) printf("%d\n", res[i]);
}
