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 maxn = 1e5 + 5;
struct node {
  int l, r, blk, id;
  friend bool operator<(node a, node b) {
    return a.blk == b.blk ? a.r < b.r : a.l < b.l;
  }
} p[maxn];
int a[maxn], cnt[maxn], ans1, ans2, n, m, mn[maxn], mx[maxn],
    l = 1, r, sn = 256, bel[maxn], sum[maxn];
vector<int> v[maxn];
deque<int> q[maxn];
bool check(int c) {
  if (q[c].size() <= 2) return 1;
  return bel[*++q[c].begin()] == bel[q[c].back()];
}
void revise(int x, bool p, bool v) {
  int c = a[x];
  v ? ans1 += !cnt[c]++ : ans1 -= !--cnt[c];
  if (!q[c].empty()) ans2 -= check(c);
  if (v)
    p ? q[c].push_back(x) : q[c].push_front(x);
  else
    p ? q[c].pop_back() : q[c].pop_front();
  if (!q[c].empty()) ans2 += check(c);
}
void md(int i) {
  while (l > p[i].l) revise(--l, 0, 1);
  while (r < p[i].r) revise(++r, 1, 1);
  while (l < p[i].l) revise(l++, 0, 0);
  while (r > p[i].r) revise(r--, 1, 0);
  sum[p[i].id] = ans1 + (ans2 == 0);
}
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; i++)
    scanf("%d", &a[i]), v[a[i]].push_back(i);
  for (register int i = 1; i <= n; i++) {
    if (v[i].size() <= 2) continue;
    for (register int j = 2; j < v[i].size(); j++)
      bel[v[i][j]] = bel[v[i][j - 1]] +
                     (v[i][j] - v[i][j - 1] != v[i][j - 1] - v[i][j - 2]);
  }
  scanf("%d", &m);
  for (register int i = 1; i <= m; i++)
    scanf("%d%d", &p[i].l, &p[i].r), p[i].blk = p[i].l / sn, p[i].id = i;
  sort(p + 1, p + m + 1);
  for (register int i = 1; i <= m; i++) md(i);
  for (register int i = 1; i <= m; i++) printf("%d\n", sum[i]);
  return 0;
}
