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 long long MAXN = 1e6 + 10;
long long N, M, val[MAXN], nxt[MAXN], lst[MAXN], pre[MAXN], fa[MAXN], ans[MAXN];
struct que {
  long long l, r, id;
  friend bool operator<(que a, que b) { return a.r < b.r; }
} Q[MAXN];
struct arr {
  long long C[MAXN];
  void add(long long x, long long v) {
    if (!x) return;
    while (x <= N) {
      C[x] += v;
      x += (x & (-x));
    }
  }
  long long ask(long long x) {
    long long ret = 0;
    while (x) {
      ret += C[x];
      x -= (x & (-x));
    }
    return ret;
  }
} t1, t2;
int main() {
  scanf("%lld", &N);
  for (long long i = 1; i <= N; i++) {
    scanf("%lld", val + i);
    nxt[lst[val[i]]] = i;
    pre[i] = lst[val[i]];
    lst[val[i]] = i;
    if (i - pre[i] == pre[i] - pre[pre[i]])
      fa[i] = fa[pre[i]];
    else
      fa[i] = pre[pre[i]];
  }
  scanf("%lld", &M);
  for (long long i = 1; i <= M; i++)
    scanf("%lld%lld", &Q[i].l, &Q[i].r), Q[i].id = i;
  sort(Q + 1, Q + M + 1);
  long long now = 0;
  for (long long i = 1; i <= M; i++) {
    while (now < Q[i].r) {
      ++now;
      t1.add(now, 1);
      t1.add(pre[now], -1);
      t2.add(fa[now], 1);
      t2.add(fa[pre[now]], -1);
    }
    long long tem1 = t1.ask(Q[i].r) - t1.ask(Q[i].l - 1),
              tem2 = t2.ask(Q[i].r) - t2.ask(Q[i].l - 1);
    ans[Q[i].id] = tem1 + (tem1 == tem2);
  }
  for (long long i = 1; i <= M; i++) printf("%lld\n", ans[i]);
  return 0;
}
