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 = 100 * 1000 + 5;
int n, q;
int a[N];
int b[N];
int lt[N];
int prv[N];
int seg1[N * 4];
int seg2[N * 4];
vector<pair<int, int> > vec[N];
int ans[N];
void update(int p, int x, int id = 1, int s = 1, int e = n + 1) {
  if (p < s || e <= p) {
    return;
  }
  if (e - s == 1) {
    seg1[id] = x;
    seg2[id] = b[p];
    return;
  }
  int md = (s + e) / 2;
  update(p, x, id * 2, s, md);
  update(p, x, id * 2 + 1, md, e);
  seg1[id] = seg1[id * 2] + seg1[id * 2 + 1];
  seg2[id] = min(seg2[id * 2], seg2[id * 2 + 1]);
}
pair<int, int> get(int l, int r, int id = 1, int s = 1, int e = n + 1) {
  if (e <= l || r <= s) {
    return {0, N};
  }
  if (l <= s && e <= r) {
    return {seg1[id], seg2[id]};
  }
  int md = (s + e) / 2;
  pair<int, int> p1 = get(l, r, id * 2, s, md),
                 p2 = get(l, r, id * 2 + 1, md, e), p;
  p.first = p1.first + p2.first;
  p.second = min(p1.second, p2.second);
  return p;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    prv[i] = lt[a[i]];
    lt[a[i]] = i;
  }
  cin >> q;
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    vec[r].push_back({l, i});
  }
  fill(seg2, seg2 + n * 4 + 1, N);
  fill(b, b + n + 1, N);
  for (int i = 1; i <= n; i++) {
    int x = prv[i], y = 0;
    if (x != 0) {
      y = prv[x];
    }
    if (x != 0 && y != 0) {
      if (i - x == x - y) {
        b[i] = b[x];
      } else {
        b[i] = y;
      }
    } else {
      b[i] = 0;
    }
    update(i, 1);
    if (x != 0) {
      b[x] = N;
      update(prv[i], 0);
    }
    for (auto p : vec[i]) {
      int l = p.first, f = p.second;
      pair<int, int> t = get(l, i + 1);
      ans[f] = t.first;
      if (t.second >= l) {
        ans[f]++;
      }
    }
  }
  for (int i = 0; i < q; i++) {
    cout << ans[i] << endl;
  }
  return 0;
}
