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: Accepted)

#include<stdio.h>
#include<algorithm>
using namespace std;
#define MLOG 17
int n,m,q;
int a[100005];
int p[100005][18];
void init()
{
    int i,x,l,r,mid,ans;
    for(i=1;i<=n;i++)
    {
        ans = i;
        l = i; r = n;
        while(l<=r)
        {
            mid = (l+r)/2;
            if(a[mid]-a[i]<=m)
            {
                ans = mid;
                l = mid+1;
            }
            else r = mid-1;
        }
        p[i][0] = ans;
    }
    for(i=1;i<=MLOG;i++)
    {
        for(x=1;x<=n;x++) p[x][i] = p[p[x][i-1]][i-1];
    }
}
int get(int x,int cnt)
{
    int i,now;
    now = x;
    for(i=0;i<=MLOG;i++)
    {
        if(cnt&1) now = p[now][i];
        cnt/=2;
    }
    return now;
}
main()
{
    int i,x,y,l,r,mid,ans;
    scanf("%d",&n);
    for(i=1;i<=n;i++) scanf("%d",&a[i]);
    scanf("%d%d",&m,&q);
    init();
    for(i=0;i<q;i++)
    {
        scanf("%d%d",&x,&y);
        if(x>y) swap(x,y);
        l = 0; r = n;
        while(l<=r)
        {
            mid = (l+r)/2;
            if(get(x,mid)>=y)
            {
                ans = mid;
                r = mid-1;
            }
            else l = mid+1;
        }
        printf("%d\n",ans);
    }
}