Problem:
Score : 700 points

Problem Statement

N hotels are located on a straight line. The coordinate of the i-th hotel (1 \leq i \leq N) is x_i.

Tak the traveler has the following two personal principles:

He never travels a distance of more than L in a single day.

He never sleeps in the open. That is, he must stay at a hotel at the end of a day.

You are given Q queries. The j-th (1 \leq j \leq Q) query is described by two distinct integers a_j and b_j.
For each query, find the minimum number of days that Tak needs to travel from the a_j-th hotel to the b_j-th hotel following his principles.
It is guaranteed that he can always travel from the a_j-th hotel to the b_j-th hotel, in any given input.

Constraints

2 \leq N \leq 10^5

1 \leq L \leq 10^9

1 \leq Q \leq 10^5

1 \leq x_i < x_2 < ... < x_N \leq 10^9

x_{i+1} - x_i \leq L

1 \leq a_j,b_j \leq N

a_j \neq b_j

N,\,L,\,Q,\,x_i,\,a_j,\,b_j are integers.

Partial Score

200 points will be awarded for passing the test set satisfying N \leq 10^3 and Q \leq 10^3.

Input

The input is given from Standard Input in the following format:

N
x_1 x_2 ... x_N
L
Q
a_1 b_1
a_2 b_2
:
a_Q b_Q

Output

Print Q lines.
The j-th line (1 \leq j \leq Q) should contain the minimum number of days that Tak needs to travel from the a_j-th hotel to the b_j-th hotel.

Sample Input 1

9
1 3 6 13 15 18 19 29 31
10
4
1 8
7 3
6 7
8 5

Sample Output 1

4
2
1
2

For the 1-st query, he can travel from the 1-st hotel to the 8-th hotel in 4 days, as follows:

Day 1: Travel from the 1-st hotel to the 2-nd hotel. The distance traveled is 2.

Day 2: Travel from the 2-nd hotel to the 4-th hotel. The distance traveled is 10.

Day 3: Travel from the 4-th hotel to the 7-th hotel. The distance traveled is 6.

Day 4: Travel from the 7-th hotel to the 8-th hotel. The distance traveled is 10.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <cstdio>
#include <algorithm>

using namespace std;

#define REP(i,n)   for(int i=0; i<(int)(n); i++)
#define FOR(i,b,e) for(int i=(b); i<=(int)(e); i++)
#define DUMP(a, n) REP(i, n) printf("%d%c", a[i], i + 1 == n ? '\n' : ' ')
#define DUMP2D(a, n, m) REP(i, n) REP(j, m) printf("%d%c", a[i][j], j + 1 == m ? '\n' : ' '); puts("")

const int N_MAX = 100000;
const int Q_MAX = 100000;
const int LOGN_MAX = 17;

int N;
int x[N_MAX];
int L;
int Q;
int a[Q_MAX];
int b[Q_MAX];

int nxt[LOGN_MAX][N_MAX];

void solve() {
  int t = 0;
  REP(f, N) {
    nxt[0][f] = t - 1;
    while (t < N && x[t] - x[f] <= L) {
      nxt[0][f] = t++;
    }
  }
  nxt[0][N - 1] = N;

  REP(k, LOGN_MAX - 1) {
    REP(i, N) {
      int j = nxt[k][i];
      nxt[k + 1][i] = nxt[k][j] > j ? nxt[k][j] : N;
    }
  }

  // DUMP2D(nxt, LOGN_MAX, N);

  REP(i, Q) {
    int f = a[i];
    int t = b[i];
    int d = 0;
    for (int k = LOGN_MAX - 1; k >= 0; k--) {
      if (nxt[k][f] <= t) {
        d |= 1 << k;
        f = t;
      }
    }
    printf("%d\n", d);
  }
}

void input() {
  scanf("%d", &N);
  REP(i, N) scanf("%d", x + i);
  scanf("%d", &L);
  scanf("%d", &Q);
  REP(i, Q) {
    scanf("%d%d", a + i, b + i);
    --a[i];
    --b[i];
    if (a[i] > b[i]) swap(a[i], b[i]);
  }
}

int main() {
  input();
  solve();
  return 0;
}