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 = 100000 + 10;
int a[maxn];
int n;
int last[maxn];
int lf[maxn];
int dif[maxn], ext[maxn];
void init() {
  memset(last, 0, sizeof(last));
  for (int i = 1; i <= n; i++) {
    lf[i] = last[a[i]];
    last[a[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    if (lf[i] == 0)
      dif[i] = 0, ext[i] = 0;
    else if (dif[lf[i]] == i - lf[i]) {
      dif[i] = i - lf[i];
      ext[i] = ext[lf[i]];
    } else if (dif[lf[i]] == 0) {
      dif[i] = i - lf[i];
      ext[i] = 0;
    } else {
      dif[i] = i - lf[i];
      ext[i] = lf[i] - dif[lf[i]];
    }
  }
}
vector<pair<int, int>> q[maxn];
int sum[maxn];
int cnt[maxn];
void go(int i, int add) {
  while (i) {
    sum[i] += add;
    i -= i & (-i);
  }
}
void update(int st, int en) {
  go(en, 1);
  go(st - 1, -1);
}
int getsum(int i) {
  int ans = 0;
  while (i <= n) {
    ans += sum[i];
    i += i & (-i);
  }
  return ans;
}
int tree[maxn];
void go1(int i, int add) {
  while (i) {
    tree[i] += add;
    i -= i & (-i);
  }
}
void add(int st, int en) {
  go1(en, 1);
  go1(st - 1, -1);
}
void sub(int st, int en) {
  go1(en, -1);
  go1(st - 1, 1);
}
int gettree(int i) {
  int ans = 0;
  while (i <= n) {
    ans += tree[i];
    i += i & (-i);
  }
  return ans;
}
void solve() {
  memset(sum, 0, sizeof(sum));
  for (int i = 1; i <= n; i++) {
    update(lf[i] + 1, i);
    for (int j = 0; j < q[i].size(); j++) {
      pair<int, int> it = q[i][j];
      cnt[it.second] = getsum(it.first);
    }
  }
  memset(tree, 0, sizeof(tree));
  for (int i = 1; i <= n; i++) {
    if (lf[i] != 0) {
      sub(ext[lf[i]] + 1, lf[i]);
    }
    add(ext[i] + 1, i);
    for (int j = 0; j < q[i].size(); j++) {
      pair<int, int> it = q[i][j];
      int t = gettree(it.first);
      cnt[it.second] += t > 0 ? 0 : 1;
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  init();
  for (int i = 1; i <= n; i++) q[i].clear();
  int ct;
  scanf("%d", &ct);
  for (int i = 0; i < ct; i++) {
    int st, en;
    scanf("%d%d", &st, &en);
    q[en].push_back(make_pair(st, i));
  }
  solve();
  for (int i = 0; i < ct; i++) printf("%d\n", cnt[i]);
  return 0;
}
