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 segtree {
  int cnt[1024 * 128 * 2];
  int mum;
  void init(int n) {
    mum = 1;
    while (mum < n) mum <<= 1;
    for (int i = 0; i < mum * 2; i++) {
      cnt[i] = 0;
    }
  }
  void pl(int a, int b, int k, int l, int r) {
    if (b <= l || r <= a) return;
    if (a <= l && r <= b)
      cnt[k]++;
    else {
      pl(a, b, k * 2 + 1, l, (l + r) / 2);
      pl(a, b, k * 2 + 2, (l + r) / 2, r);
    }
  }
  void pl(int a, int b) { pl(a, b, 0, 0, mum); }
  int get(int k) {
    k += mum - 1;
    int c = cnt[k];
    while (k > 0) {
      k = (k - 1) / 2;
      c += cnt[k];
    }
    return c;
  }
};
struct segtree2 {
  int mx[1024 * 128 * 2];
  int mum;
  void init(int n) {
    mum = 1;
    while (mum < n) mum <<= 1;
    for (int i = 0; i < mum * 2; i++) {
      mx[i] = 0;
    }
  }
  void add(int k, int x) {
    k += mum - 1;
    mx[k] = x;
    while (k > 0) {
      k = (k - 1) / 2;
      mx[k] = max(mx[k * 2 + 1], mx[k * 2 + 2]);
    }
  }
  int get(int a, int b, int k, int l, int r) {
    if (b <= l || r <= a) return 0;
    if (a <= l && r <= b)
      return mx[k];
    else {
      int vl = get(a, b, k * 2 + 1, l, (l + r) / 2);
      int vr = get(a, b, k * 2 + 2, (l + r) / 2, r);
      return max(vl, vr);
    }
  }
  int get(int a, int b) { return get(a, b, 0, 0, mum); }
};
segtree seg;
segtree2 nn;
vector<pair<int, int> > query[100005];
int ans[100005], nxt[100005], bef[100005];
int ng[100005], diff[100005];
int in[100005];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &in[i]);
    in[i]--;
  }
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    l--;
    r--;
    query[l].push_back(pair<int, int>(r, i));
  }
  seg.init(n + 2);
  nn.init(n + 2);
  memset(bef, -1, sizeof(bef));
  for (int i = n - 1; i >= 0; i--) {
    nxt[i] = bef[in[i]];
    bef[in[i]] = i;
    if (nxt[i] == -1) {
      diff[i] = 100005;
      ng[i] = n;
    } else {
      nn.add(nxt[i], -1);
      int d = nxt[i] - i;
      if (nxt[nxt[i]] == -1) {
        diff[i] = d;
        ng[i] = n;
      } else {
        diff[i] = d;
        if (diff[nxt[i]] == d)
          ng[i] = ng[nxt[i]];
        else
          ng[i] = nxt[nxt[i]];
      }
    }
    nn.add(i, ng[i]);
    if (nxt[i] != -1)
      seg.pl(i, nxt[i]);
    else
      seg.pl(i, n);
    for (int j = 0; j < query[i].size(); j++) {
      pair<int, int> p = query[i][j];
      int to = p.first;
      int cnt = seg.get(to);
      int last = nn.get(i, to + 1);
      ans[p.second] = cnt + (to < last ? 0 : 1);
    }
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
  return 0;
}
