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;
struct query {
  int l, r, p, id;
  bool operator<(const query &b) const {
    return p < b.p || (p == b.p && r < b.r);
  }
} q[100001];
int n, m, a[100001], cnt[100001], d[100001], ans[100001], sum, sumd;
deque<int> v[100001];
unordered_map<int, int> k[100001];
void push(int x, bool front) {
  if (!cnt[a[x]]++) sumd++;
  int p1, p2;
  if (front)
    p1 = 1, p2 = 0, v[a[x]].push_front(x);
  else
    p1 = v[a[x]].size(), p2 = p1 - 1, v[a[x]].push_back(x);
  if (v[a[x]].size() > 1 && !k[a[x]][v[a[x]][p1] - v[a[x]][p2]]++ &&
      k[a[x]].size() == 2)
    sum++;
}
void pop(int x, bool front) {
  if (!--cnt[a[x]]) sumd--;
  int p1, p2;
  if (front)
    p1 = 1, p2 = 0;
  else
    p1 = v[a[x]].size() - 1, p2 = p1 - 1;
  if (v[a[x]].size() > 1 && !--k[a[x]][v[a[x]][p1] - v[a[x]][p2]]) {
    if (k[a[x]].size() == 2) sum--;
    k[a[x]].erase(v[a[x]][p1] - v[a[x]][p2]);
  }
  if (front)
    v[a[x]].pop_front();
  else
    v[a[x]].pop_back();
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  const int z = sqrt(n) + .5;
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &q[i].l, &q[i].r);
    q[i].p = q[i].l / z;
    q[i].id = i;
  }
  sort(q + 1, q + 1 + m);
  int l = 1, r = 0;
  for (int i = 1; i <= m; i++) {
    while (l > q[i].l) push(--l, 1);
    while (r < q[i].r) push(++r, 0);
    while (l < q[i].l) pop(l++, 1);
    while (r > q[i].r) pop(r--, 0);
    ans[q[i].id] = sumd + (sum == sumd);
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}
