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 = 2e5 + 10;
int n, t, sq = 300, k;
int a[MAXN];
long long ans[MAXN];
int l[MAXN], r[MAXN], id[MAXN];
deque<int> dq[MAXN], dq2[MAXN];
bool ful[MAXN], ok[MAXN], na[MAXN];
int res, res2;
bool cmp(int id1, int id2) {
  if (l[id1] / sq != l[id2] / sq) return l[id1] / sq < l[id2] / sq;
  if ((l[id1] / sq) % 2) return r[id2] < r[id1];
  return r[id1] < r[id2];
}
void upd(int x) {
  if (dq[x].size() && !ful[x]) ful[x] = true, res2++;
  if (!dq[x].size() && ful[x]) ful[x] = false, res2--;
  if (dq2[x].size() && !na[x]) na[x] = true, res++;
  if (!dq2[x].size() && na[x]) na[x] = false, res--;
}
void del(int ind, bool bl) {
  if (bl) {
    dq[a[ind]].pop_back();
    if (dq2[a[ind]].size() && dq[a[ind]].back() == dq2[a[ind]].back())
      dq2[a[ind]].pop_back();
  } else {
    dq[a[ind]].pop_front();
    if (dq2[a[ind]].size() && dq2[a[ind]].front() == dq[a[ind]].front())
      dq2[a[ind]].pop_front();
  }
  upd(a[ind]);
}
void add(int ind, bool bl) {
  if (bl) {
    int val = a[ind];
    dq[val].push_back(ind);
    int x = dq[val].size();
    if (dq[val].size() > 2 &&
        dq[val][x - 1] - dq[val][x - 2] != dq[val][x - 2] - dq[val][x - 3])
      dq2[val].push_back(dq[val][x - 2]);
  } else {
    int val = a[ind];
    dq[val].push_front(ind);
    if (dq[val].size() > 2 &&
        dq[val][0] - dq[val][1] != dq[val][1] - dq[val][2])
      dq2[val].push_front(dq[val][1]);
  }
  upd(a[ind]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  cin >> t;
  for (int i = 0; i < t; i++) cin >> l[i] >> r[i], id[i] = i, l[i]--;
  sort(id, id + t, cmp);
  int L = 0, R = 0;
  for (int i = 0; i < t; i++) {
    while (L > l[id[i]]) {
      L--;
      add(L, 0);
    }
    while (R < r[id[i]]) {
      add(R, 1);
      R++;
    }
    while (L < l[id[i]]) {
      del(L, 0);
      L++;
    }
    while (R > r[id[i]]) {
      R--;
      del(R, 1);
    }
    ans[id[i]] = res2 + (res2 == res);
  }
  for (int i = 0; i < t; i++) cout << ans[i] << '\n';
  return 0;
}
