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.

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.

Example

Input

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


Output

4
2
1
2
#include<bits/stdc++.h>
#define N 100010
using namespace std;
template<typename T>inline void read(T &x)
{
	x=0;
	static int p;p=1;
	static char c;c=getchar();
	while(!isdigit(c)){if(c=='-')p=-1;c=getchar();}
	while(isdigit(c)) {x=(x<<1)+(x<<3)+(c-'0');c=getchar();}
	x*=p;
}
int n;
int h[N],l,q;
int st[N][19];
int main()
{
	read(n);
	for(int i=1;i<=n;i++) read(h[i]);
	read(l);read(q);
	for(int i=1;i<=n;i++)
	{
		int pos=upper_bound(h+1,h+n+1,h[i]+l)-h-1;
		if(h[i]+l>=h[n])st[i][0]=n;
		else st[i][0]=pos;
	}
	for(int len=1;len<=18;len++)
		for(int i=1;i<=n;i++) st[i][len]=st[st[i][len-1]][len-1];
	while(q--)
	{
		static int a,b;
		read(a);read(b);
		if(a>b)swap(a,b);
		int now=a,ans=0;
		for(int i=18;i>=0;i--)
			if(st[now][i]<b)now=st[now][i],ans+=(1<<i);
		printf("%d\n",ans+1);
	}
	return 0;
}