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>
using namespace std;
int n,l,m,a[100003],x,y,b[100003][103],ans,l1,r1,mid;
int main()
{
	cin>>n;
	for(int i=1; i<=n; i++)
		scanf("%d",&a[i]);
	cin>>l>>m;
	for(int i=1; i<=n-1; i++)
	{
		l1=i+1,r1=n;
		while(l1<=r1)
		{
			mid=(l1+r1)/2;
			if(a[mid]-a[i]>l)
				r1=mid-1;
			else
				l1=mid+1;
		}
		b[i][0]=r1;
	}
	for(int j=n; j>=1; j--)
		for(int i=1; b[b[j][i-1]][i-1]!=0; i++)
			b[j][i]=b[b[j][i-1]][i-1];
	for(int i=1; i<=m; i++)
	{
		scanf("%d%d",&x,&y);
		ans=0;
		if(x>y)
			swap(x,y);
		while(x<y)
		{
			for(int i=0; i==0||b[x][i-1]!=0; i++)
				if(b[x][i]==y)
				{
					x=b[x][i],ans+=(1<<i);
					break;
				}
				else if(b[x][i]>y||b[x][i]==0)
				{
					if(i==0)
					{
						x=y,ans+=1;
						break;
					}
					x=b[x][i-1],ans+=(1<<(i-1));
					break;
				}
		}
		cout<<ans<<endl;
	}
	return 0;
}