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 int N = 1e5 + 10, SQ = 450, inf = 1e5 + 10;
long long int n, m, q, l, r, a[N], f[N], fa[N], b[N], c[N], res[N], vq[10000];
pair<long long int, long long int> p, g[N];
vector<pair<long long int, long long int> > v[N];
void add(long long int v, long long int c) {
  for (; v < N; v += v & -v) fa[v] += c;
}
long long int gets(long long int ind) {
  long long int ret = 0;
  for (; ind; ind -= ind & -ind) ret += fa[ind];
  return ret;
}
void upd(long long int s) {
  long long int id = s / SQ;
  vq[id] = inf;
  for (int i = id * SQ; i < (id + 1) * SQ && i < n; i++)
    vq[id] = min(vq[id], c[i]);
}
long long int get(long long int s, long long int e) {
  long long int l = s / SQ, r = e / SQ, ret = inf;
  l++;
  for (int i = l; i < r; i++) ret = min(ret, vq[i]);
  if (l - r == 1) {
    for (int i = s; i <= e; i++) ret = min(ret, c[i]);
    return ret;
  }
  for (int i = s; i < l * SQ; i++) ret = min(ret, c[i]);
  for (int i = r * SQ; i <= e; i++) ret = min(ret, c[i]);
  return ret;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  fill(b, b + N, -1);
  cin >> q;
  for (int i = 1; i <= q; i++) {
    cin >> l >> r;
    v[r - 1].push_back({l - 1, i});
  }
  for (int i = 0; i < n; i++) {
    f[i] = b[a[i]];
    b[a[i]] = i;
    add(i + 1, 1);
    g[i] = make_pair(-1, 0);
    if (f[i] != -1) {
      add(f[i] + 1, -1);
      if (g[f[i]].second == 0 || g[f[i]].second != i - f[i])
        g[i] = make_pair(f[f[i]], i - f[i]);
      else if (g[f[i]].second == i - f[i])
        g[i] = g[f[i]];
      c[f[i]] = inf;
      upd(f[i]);
    }
    c[i] = g[i].first;
    upd(i);
    r = i;
    for (auto y : v[i]) {
      l = y.first;
      res[y.second] = gets(r + 1) - gets(l);
      if (get(l, r) >= l) res[y.second]++;
    }
  }
  for (int i = 1; i <= q; i++) cout << res[i] << '\n';
}
