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;
long long cnt[100005], a[100005], sum[100005], ans[100005],
    answer = 0, n, BLOCK, i, currentL, currentR, L, R;
deque<long long> qu[100005];
set<long long> s;
struct node {
  long long L, R, i;
} q[100005];
bool cmp(node x, node y) {
  if (x.L / BLOCK != y.L / BLOCK) {
    return x.L / BLOCK < y.L / BLOCK;
  }
  return x.R < y.R;
}
int main() {
  BLOCK = ceil(sqrt(100005));
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  int m;
  scanf("%d", &m);
  for (i = 0; i < m; i++) {
    scanf("%d%d", &q[i].L, &q[i].R);
    q[i].L--;
    q[i].R--;
    q[i].i = i;
  }
  sort(q, q + m, cmp);
  answer = 1;
  cnt[a[0]] = 1;
  s.insert(a[0]);
  qu[a[0]].push_back(0);
  currentL = 0, currentR = 0;
  for (i = 0; i < m; i++) {
    long long L = q[i].L, R = q[i].R;
    while (currentL > L) {
      currentL--;
      cnt[a[currentL]]++;
      if (cnt[a[currentL]] == 1) {
        answer++;
      }
      deque<long long>& v = qu[a[currentL]];
      v.push_front(currentL);
      if (v.size() >= 3 && v[0] - v[1] != v[1] - v[2]) {
        sum[a[currentL]]++;
      }
      if (!sum[a[currentL]] && cnt[a[currentL]]) {
        s.insert(a[currentL]);
      } else {
        if (s.find(a[currentL]) != s.end()) {
          s.erase(a[currentL]);
        }
      }
    }
    while (currentR < R) {
      currentR++;
      cnt[a[currentR]]++;
      if (cnt[a[currentR]] == 1) {
        answer++;
      }
      deque<long long>& v = qu[a[currentR]];
      v.push_back(currentR);
      long long z = v.size();
      if (v.size() >= 3 && v[z - 1] - v[z - 2] != v[z - 2] - v[z - 3]) {
        sum[a[currentR]]++;
      }
      if (!sum[a[currentR]] && cnt[a[currentR]]) {
        s.insert(a[currentR]);
      } else {
        if (s.find(a[currentR]) != s.end()) {
          s.erase(a[currentR]);
        }
      }
    }
    while (currentR > R) {
      cnt[a[currentR]]--;
      if (cnt[a[currentR]] == 0) {
        answer--;
      }
      deque<long long>& v = qu[a[currentR]];
      long long z = v.size();
      if (v.size() >= 3 && v[z - 1] - v[z - 2] != v[z - 2] - v[z - 3]) {
        sum[a[currentR]]--;
      }
      if (!sum[a[currentR]] && cnt[a[currentR]]) {
        s.insert(a[currentR]);
      } else {
        if (s.find(a[currentR]) != s.end()) {
          s.erase(a[currentR]);
        }
      }
      v.pop_back();
      currentR--;
    }
    while (currentL < L) {
      cnt[a[currentL]]--;
      if (cnt[a[currentL]] == 0) {
        answer--;
      }
      deque<long long>& v = qu[a[currentL]];
      if (v.size() >= 3 && v[0] - v[1] != v[1] - v[2]) {
        sum[a[currentL]]--;
      }
      if (!sum[a[currentL]] && cnt[a[currentL]]) {
        s.insert(a[currentL]);
      } else {
        if (s.find(a[currentL]) != s.end()) {
          s.erase(a[currentL]);
        }
      }
      v.pop_front();
      currentL++;
    }
    if (L == 1 && R == 9) {
    }
    ans[q[i].i] = answer + 1;
    if (s.size()) {
      ans[q[i].i]--;
    }
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
}
