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 number {
  deque<int> q;
  map<int, int> db;
  int curr = 0;
};
struct pii {
  int x, y, id;
};
const int S = 300;
bool sortFunc(pii a, pii b) {
  if (a.x / S == b.x / S) return ((a.x / S) & 1 ? a.y > b.y : a.y < b.y);
  return a.x < b.x;
}
vector<number> t(1e5 + 1);
void inc(map<int, int>& db, int num, int& curr) {
  if (db.count(num) == 0) db[num] == 0;
  if (++db[num] == 1) curr++;
}
void dec(map<int, int>& db, int num, int& curr) {
  if (--db[num] == 0) curr--;
}
int curr = 0, nums = 0;
void add(int poz, int num) {
  if (t[num].curr <= 1 && t[num].q.size() > 0) curr--;
  if (t[num].q.empty())
    t[num].q.push_front(poz);
  else if (poz < t[num].q.front()) {
    inc(t[num].db, t[num].q.front() - poz, t[num].curr);
    t[num].q.push_front(poz);
  } else {
    inc(t[num].db, poz - t[num].q.back(), t[num].curr);
    t[num].q.push_back(poz);
  }
  if (t[num].curr <= 1) curr++;
  if (t[num].q.size() == 1) nums++;
}
void remove(int poz, int num) {
  if (t[num].curr <= 1) curr--;
  if (t[num].q.size() == 1)
    t[num].q.pop_front();
  else if (poz == t[num].q.front()) {
    t[num].q.pop_front();
    dec(t[num].db, t[num].q.front() - poz, t[num].curr);
  } else {
    t[num].q.pop_back();
    dec(t[num].db, poz - t[num].q.back(), t[num].curr);
  }
  if (t[num].curr <= 1 && t[num].q.size() > 0) curr++;
  if (t[num].q.size() == 0) nums--;
}
int main() {
  int test;
  int n;
  cin >> n;
  vector<int> sor(n);
  for (int& a : sor) cin >> a;
  int q;
  cin >> q;
  vector<pii> v(q);
  for (int i = 0; i < q; i++)
    cin >> v[i].x >> v[i].y, v[i].id = i, v[i].x--, v[i].y--;
  sort(v.begin(), v.end(), sortFunc);
  int l = 0, r = -1;
  vector<int> ans(q);
  for (pii& a : v) {
    while (a.x < l) add(l - 1, sor[l - 1]), l--;
    while (r < a.y) add(r + 1, sor[r + 1]), r++;
    while (l < a.x) remove(l, sor[l]), l++;
    while (a.y < r) remove(r, sor[r]), r--;
    ans[a.id] = (curr > 0 ? nums : nums + 1);
  }
  for (int& a : ans) cout << a << "\n";
}
