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 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
int a[100005];
int dp[100005][50];
int main(){
    int n;
    scanf("%d",&n);
    for(int i = 1;i <= n;i++)
        scanf("%d",&a[i]);
    int L,Q;
    scanf("%d%d",&L,&Q);
    for(int i = 1;i <= n;i++){
        int id = upper_bound(a+1,a+1+n,a[i]+L)-a-1;
        if(a[i] + L >= a[n])
            dp[i][0] = n;
        else dp[i][0] = id;}
    for(int i = 1;i <= 30;i++){
        for(int j = 1;j <= n;j++)
            dp[j][i] = dp[dp[j][i-1]][i-1];}
    while(Q--){
        int L,R;
        scanf("%d%d",&L,&R);
        if(R<L)
            swap(L,R);  
        long long ans=0;
        for(int i = 30;i >= 0;i--){
            if(dp[L][i] < R){
                ans=ans+(1ll<<i);
                L=dp[L][i];}}printf("%lld\n",ans+1);}
    return 0;
}