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 INF = 2147483647;
template <class T>
int size(const T &x) {
  return x.size();
}
list<int> inst[100100];
int arr[100100], res[100100], sum[100100], inside = 0, ok = 0, st;
bool cmp(const pair<pair<int, int>, int> &a,
         const pair<pair<int, int>, int> &b) {
  return pair<int, int>(a.first.first / st, a.first.second) <
         pair<int, int>(b.first.first / st, b.first.second);
}
void add(int i, bool left) {
  int cur = arr[i];
  int add = 0;
  if (inst[cur].empty()) {
    ok++;
    inside++;
  }
  if (left) {
    inst[cur].push_front(i);
    list<int>::const_iterator it = inst[cur].begin();
    int a = *it;
    it++;
    if (it != inst[cur].end()) {
      int b = *it;
      it++;
      if (it != inst[cur].end()) {
        int c = *it;
        add += abs((b - a) - (c - b));
      }
    }
  } else {
    inst[cur].push_back(i);
    list<int>::reverse_iterator it = inst[cur].rbegin();
    int c = *it;
    it++;
    if (it != inst[cur].rend()) {
      int b = *it;
      it++;
      if (it != inst[cur].rend()) {
        int a = *it;
        add += abs((b - a) - (c - b));
      }
    }
  }
  if (add > 0 && sum[cur] == 0) {
    ok--;
  }
  sum[cur] += add;
}
void remove(int i, bool left) {
  int cur = arr[i];
  int add = 0;
  assert(!inst[cur].empty());
  if (left) {
    list<int>::const_iterator it = inst[cur].begin();
    int a = *it;
    it++;
    if (it != inst[cur].end()) {
      int b = *it;
      it++;
      if (it != inst[cur].end()) {
        int c = *it;
        add -= abs((b - a) - (c - b));
      }
    }
    inst[cur].pop_front();
  } else {
    list<int>::reverse_iterator it = inst[cur].rbegin();
    int c = *it;
    it++;
    if (it != inst[cur].rend()) {
      int b = *it;
      it++;
      if (it != inst[cur].rend()) {
        int a = *it;
        add -= abs((b - a) - (c - b));
      }
    }
    inst[cur].pop_back();
  }
  if (sum[cur] != 0 && sum[cur] + add == 0) {
    ok++;
  }
  sum[cur] += add;
  if (inst[cur].empty()) {
    ok--;
    inside--;
  }
}
int main() {
  memset(sum, 0, sizeof(sum));
  int n;
  scanf("%d\n", &n);
  st = static_cast<int>(sqrt(n) + 1);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
  }
  int m;
  scanf("%d\n", &m);
  vector<pair<pair<int, int>, int> > queries;
  for (int i = 0; i < m; i++) {
    int l, r;
    scanf("%d %d\n", &l, &r);
    l--, r--;
    queries.push_back(make_pair(pair<int, int>(l, r), i));
  }
  sort((queries).begin(), (queries).end(), cmp);
  int l = 0, r = 0;
  add(0, true);
  for (int i = 0; i < size(queries); i++) {
    while (l > queries[i].first.first) add(--l, true);
    while (r < queries[i].first.second) add(++r, false);
    while (l < queries[i].first.first) remove(l++, true);
    while (r > queries[i].first.second) remove(r--, false);
    res[queries[i].second] = inside;
    if (ok == 0) res[queries[i].second]++;
  }
  for (int i = 0; i < m; i++) {
    cout << res[i] << endl;
  }
  return 0;
}
