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 N = 1e5 + 5;
namespace SegTree {
int key[N << 2];
inline void Clear() { memset(key, 0, sizeof key); }
inline void Add(int l, int r, int v, int x = 1, int L = 1, int R = 100000) {
  if (l <= L && R <= r) {
    key[x] += v;
    return;
  }
  if (l <= ((L + R) >> 1)) Add(l, r, v, x << 1, L, ((L + R) >> 1));
  if (r > ((L + R) >> 1)) Add(l, r, v, x << 1 | 1, ((L + R) >> 1) + 1, R);
}
inline int Query(int p, int x = 1, int L = 1, int R = 100000) {
  if (L == R) return key[x];
  if (p <= ((L + R) >> 1))
    return key[x] + Query(p, x << 1, L, ((L + R) >> 1));
  else
    return key[x] + Query(p, x << 1 | 1, ((L + R) >> 1) + 1, R);
}
}  // namespace SegTree
struct edge {
  int nxt, to;
} e[N];
int fir[N], cnt = 0;
int n, a[N], m;
int pre[N], delta[N], st[N], lst[N];
int ans[N], l[N], r[N];
inline void addedge(int x, int y) {
  e[++cnt] = (edge){fir[x], y};
  fir[x] = cnt;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &l[i], &r[i]);
    addedge(r[i], i);
  }
  for (int i = 1; i <= n; i++) {
    SegTree ::Add(pre[a[i]] + 1, i, 1);
    pre[a[i]] = i;
    for (int j = fir[i]; j; j = e[j].nxt)
      ans[e[j].to] = SegTree ::Query(l[e[j].to]);
  }
  SegTree ::Clear();
  memset(pre, 0, sizeof pre);
  for (int i = 1, A; i <= n; i++) {
    A = a[i];
    lst[i] = pre[a[i]];
    if (delta[A] && i - pre[A] != delta[A]) {
      SegTree ::Add(st[A] + 1, pre[A], -1);
      st[A] = lst[pre[A]];
      delta[A] = i - pre[A];
      SegTree ::Add(lst[pre[A]] + 1, i, 1);
    } else {
      SegTree ::Add(pre[A] + 1, i, 1);
      if (pre[A]) delta[A] = i - pre[A];
    }
    pre[A] = i;
    for (int j = fir[i]; j; j = e[j].nxt)
      ans[e[j].to] += (SegTree ::Query(l[e[j].to]) == 0);
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}
