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;
using pii = pair<int, int>;
using vpii = vector<pii>;
const int mxn = 100006;
struct PST {
  int val;
  PST *L, *R;
  PST() {}
  PST(int val) : val(val), L(MEM), R(MEM) {}
  PST(PST *L, PST *R) : val(L->val + R->val), L(L), R(R) {}
  static PST MEM[];
  static int SIZE;
  int get(int l, int r, int sl, int sr) {
    if (sl <= l && r <= sr) return val;
    if (sr <= ((l + r) >> 1)) return L->get(l, ((l + r) >> 1), sl, sr);
    if (((l + r) >> 1) < sl) return R->get(((l + r) >> 1) + 1, r, sl, sr);
    return L->get(l, ((l + r) >> 1), sl, sr) +
           R->get(((l + r) >> 1) + 1, r, sl, sr);
  }
  PST *set(int l, int r, int x, int ope) {
    if (l == r) return (new (MEM + ++SIZE) PST(val + ope));
    if (x <= ((l + r) >> 1))
      return (new (MEM + ++SIZE) PST(L->set(l, ((l + r) >> 1), x, ope), R));
    else
      return (new (MEM + ++SIZE) PST(L, R->set(((l + r) >> 1) + 1, r, x, ope)));
  }
};
PST PST::MEM[mxn * 20 * 5] = {{0}};
int PST::SIZE = 0;
int n;
int a[mxn];
PST *Q[mxn];
vpii P[mxn + 1];
PST *Q2[mxn];
void EXEC() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  Q[0] = PST::MEM;
  map<int, int> last;
  for (int i = 1; i <= n; ++i) {
    Q[i] = Q[i - 1]->set(0, n, last[a[i]], +1);
    last[a[i]] = i;
  }
  map<int, deque<int>> col;
  for (int i = 1; i <= n; ++i) col[a[i]].push_back(i);
  auto addRange = [](int Ll, int Lr, int Rl, int Rr) {
    P[Ll].push_back({Rl, +1});
    P[Ll].push_back({Rr + 1, -1});
    P[Lr + 1].push_back({Rl, -1});
    P[Lr + 1].push_back({Rr + 1, +1});
  };
  for (auto &value : col) {
    auto &vec = value.second;
    int L = 0;
    vec.push_front(0);
    vec.push_front(0);
    vec.push_back(n + 1);
    for (int i = 2; i + 1 < vec.size(); ++i) {
      if (vec[i] - vec[i - 1] != vec[i - 1] - vec[i - 2]) L = vec[i - 2];
      addRange(L + 1, vec[i], vec[i], vec[i + 1] - 1);
    }
  }
  Q2[0] = PST::MEM;
  for (int i = 1; i <= n; ++i) {
    Q2[i] = Q2[i - 1];
    for (auto &j : P[i])
      if (j.first <= n) Q2[i] = Q2[i]->set(0, n, j.first, j.second);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int L, R;
    scanf("%d%d", &L, &R);
    int color = Q[R]->get(0, n, 0, L - 1) - (L - 1);
    int first = !Q2[L]->get(0, n, 0, R);
    printf("%d\n", color + first);
  }
}
int main() {
  EXEC();
  return 0;
}
