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<map>
#include<iostream>
#include<iomanip>
//#include<conio.h>
#include<algorithm>
#include<vector>
#include<set>
#include<cmath>
#include<stdio.h>
#include<fstream>
#include<assert.h>
#include<time.h>
#include<queue>
#include<deque>
#include<stack>
#include<list>
//#include<windows.h>
#include<sstream>
#include<string.h>
#define mp make_pair
#define ll long long
#define all(v) v.begin(),v.end()
#define memset(a,b) memset(a,b,sizeof(a))
using namespace std;
inline void write(int x){
    if(x<0) putchar('-'),x=-x;
    if(x>9) write(x/10);
    putchar(x%10+'0');
}
inline void rd(int &val){
    int x=0;
    int f=1;
    char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
	if(ch=='-'){
		f=-1;
		ch=getchar();
	}
    while(ch>='0'&&ch<='9'){
        x=x*10+ch-'0';
        ch=getchar();
    }
    val=x*f;

}
inline void print(int x){
	write(x);
	putchar('\n');
}
int a[400005];
int ri[21][600005];
int n,l,ta,x,y;
int rem[22];
int main(){
	ios_base::sync_with_stdio(false);
	int i,j;
    #ifndef LOCAL
    	cout<<"FAsd\b";
    #endif
    rd(n);
    for(i=0;i<22;i++) rem[i]=pow(2,i);
    for(i=1;i<=n;++i) rd(a[i]);
	rd(l);
	for(i=1;i<=n;++i){
    	ri[0][i]=lower_bound(a+1,a+n+1,a[i]+l+1)-a-1;
//    	if(ri[1][i]==n+1) ri[1][i]=n+1;
	}
	for(j=1;j<20;++j) for(i=1;i<=n;i++) ri[j][i]=ri[j-1][ri[j-1][i]];
	rd(ta);
	while(ta--){
		rd(x),rd(y);
		if(x>y) swap(x,y);
		int day=2147483647;
		int l=0,r=2<<20,md;
		while(l<r-1){
			md=l+r>>1;
			int tx=x,cnt=0;
			for(i=0;i<20;i++) if(md&rem[i]) tx=ri[i][tx],cnt+=rem[i];
			if(tx<y){
				l=md;
			}else{
				r=md;
				day=min(day,cnt);
			}
		} 
		print(day);
	}
}