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 (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
int x_go_max[17][100000];//2^17日>10^5日あれば，端の町から端の町まで移動できる
int a[100000], b[100000];
int main(void){
    int n, i, j;
    scanf("%d", &n);
    int x[100000];
    for(i = 0; i < n; i++){
        scanf("%d", &x[i]);
    }
    int l;
    scanf("%d", &l);
    int q;
    scanf("%d", &q);
    
    for(i = 0; i < q; i++){
        scanf("%d%d", &a[i], &b[i]);
        a[i]--;
        b[i]--;
    }
    
    for(i = 0; i < n; i++){
        //2分探索で，最大どこからどこまで移動できるかを計算
        if(x[n-1] - x[i] <= l){
            x_go_max[0][i] = n-1;
            continue;
        }
        int left = i+1;
        int right = n;
        while(right - left > 1){
            int mid = (right - left)/2 + left;
            if(x[mid] - x[i] <= l){
                left = mid;
            }else{
                right = mid;
            }
        }
        x_go_max[0][i] = left;
    }

    for(i = 1; i < 17; i++){
        for(j = 0; j < n; j++){
            x_go_max[i][j] = x_go_max[i-1][x_go_max[i-1][j]];
        } 
    }

    for(i = 0; i < q; i++){
        int ans = 0;
        int c = a[i] > b[i] ? b[i] : a[i];
        int d = a[i] > b[i] ? a[i] : b[i];
        //cからd(c < d)に移動することを考える
        for(j = 16; j >= 0; j--){
            if(x_go_max[j][c] < d){//(1 << j)日で移動できない場合
                c = x_go_max[j][c];
                ans += 1 << j;
            }
        }
        printf("%d\n", ans+1);
    }
    return 0;
}