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 INF = 0x3f3f3f3f;
const long long INFLL = 0x3f3f3f3f3f3f3f3f;
const int MAXINT = 2147483647;
const int MAXN = 1000005;
struct Query {
  int l, r, id;
} que[MAXN];
int n, q, a[MAXN], last[MAXN], lastnot[MAXN], c1[MAXN], c2[MAXN], ans[MAXN],
    mp[MAXN];
bool cmp(Query x, Query y) { return x.r < y.r; }
int lowbit(int x) { return x & -x; }
void add1(int x, int y) {
  if (x <= 0) return;
  while (x <= n) {
    c1[x] += y;
    x += lowbit(x);
  }
  return;
}
void add2(int x, int y) {
  if (x <= 0) return;
  while (x <= n) {
    c2[x] += y;
    x += lowbit(x);
  }
  return;
}
int pre1(int x) {
  int s = 0;
  while (x >= 1) {
    s += c1[x];
    x -= lowbit(x);
  }
  return s;
}
int pre2(int x) {
  int s = 0;
  while (x >= 1) {
    s += c2[x];
    x -= lowbit(x);
  }
  return s;
}
int sum1(int x, int y) { return pre1(y) - pre1(x - 1); }
int sum2(int x, int y) { return pre2(y) - pre2(x - 1); }
int main() {
  int i, j;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    last[i] = mp[a[i]];
    mp[a[i]] = i;
    lastnot[i] = (i - last[i] == last[i] - last[last[i]]) ? lastnot[last[i]]
                                                          : last[last[i]];
  }
  scanf("%d", &q);
  for (i = 1; i <= q; i++) {
    scanf("%d%d", &que[i].l, &que[i].r);
    que[i].id = i;
  }
  sort(que + 1, que + q + 1, cmp);
  j = 0;
  for (i = 1; i <= q; i++) {
    while (j < que[i].r) {
      j++;
      add1(j, 1);
      add1(last[j], -1);
      add2(lastnot[j], 1);
      add2(lastnot[last[j]], -1);
    }
    int s1 = sum1(que[i].l, que[i].r), s2 = sum2(que[i].l, que[i].r);
    ans[que[i].id] = s1 + (s1 == s2);
  }
  for (i = 1; i <= q; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
