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 SQ = 300;
struct Query {
  int l, r, id;
  bool operator<(const Query& rhs) const {
    return l / SQ < rhs.l / SQ || (l / SQ == rhs.l / SQ && r < rhs.r);
  }
} qs[100005];
int ans[100005];
int v[100005];
int fails[100005];
deque<int> dq[100005];
bool ok(int y, int i) { return dq[y][i] * 2 == dq[y][i + 1] + dq[y][i - 1]; }
int main() {
  int n = ({
    int x;
    scanf("%d", &x);
    x;
  });
  for (int i = 0; i < n; i++) {
    v[i] = ({
      int x;
      scanf("%d", &x);
      x;
    });
  }
  int q = ({
    int x;
    scanf("%d", &x);
    x;
  });
  for (int i = 0; i < q; i++) {
    qs[i] = {({
               int x;
               scanf("%d", &x);
               x;
             }) - 1,
             ({
               int x;
               scanf("%d", &x);
               x;
             }) - 1,
             i};
  }
  sort(qs, qs + q);
  int diff = 0;
  int mov = 0;
  auto add = [&](int x) {
    int y = v[x];
    if (dq[y].empty()) diff++, mov++;
    if (!dq[y].empty() && dq[y][0] > x) {
      dq[y].push_front(x);
      if (dq[y].size() > 2 && !ok(y, 1)) {
        if (!fails[y]) mov--;
        fails[y]++;
      }
    } else {
      dq[y].push_back(x);
      if (dq[y].size() > 2 && !ok(y, dq[y].size() - 2)) {
        if (!fails[y]) mov--;
        fails[y]++;
      }
    }
  };
  auto remove = [&](int x) {
    int y = v[x];
    if (dq[y].size() == 1) mov--, diff--;
    if (dq[y][0] == x) {
      if (dq[y].size() > 2 && !ok(y, 1)) {
        fails[y]--;
        if (!fails[y]) mov++;
      }
      dq[y].pop_front();
    } else {
      if (dq[y].size() > 2 && !ok(y, dq[y].size() - 2)) {
        fails[y]--;
        if (!fails[y]) mov++;
      }
      dq[y].pop_back();
    }
  };
  int L = 0, R = -1;
  for (int k = 0; k < q; k++) {
    while (R < qs[k].r) add(++R);
    while (R > qs[k].r) remove(R--);
    while (L < qs[k].l) remove(L++);
    while (L > qs[k].l) add(--L);
    ans[qs[k].id] = diff + (mov == 0);
  }
  for (int i = 0; i < q; i++) {
    printf("%d\n", ans[i]);
  }
}
