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 MAXN 100000+5
using namespace std;
typedef long long LL;
LL n,a[MAXN],L,t,lgn;
LL ne[MAXN];
LL f[MAXN][21];
LL divide(LL d)
{
	LL l=1,r=n;
	while(l<=r)
	{
		LL mid=(l+r)>>1;
		if(a[mid]>d) r=mid-1;
		if(a[mid]<=d) l=mid+1;
	}
	return r;
}
int main()
{
	scanf("%lld",&n);
	lgn=log2(n);
	for(int i=1;i<=n;i++)
	{
		scanf("%lld",&a[i]);
	}
	scanf("%lld%lld",&L,&t);
	for(int i=1;i<=n;i++)
	{
		LL x=divide(a[i]+L);
	    f[i][0]=x;
	}
	for(int j=1;j<=lgn;j++)
	{
		for(int i=1;i<=n;i++)
		{
			f[i][j]=f[f[i][j-1]][j-1];
		}
	}
	for(int i=1;i<=t;i++)
	{
		LL times=0;
		LL u,v;
		scanf("%lld%lld",&u,&v);
		if(u>v) swap(u,v);
		for(int j=lgn;j>=0;j--)
		{
			if(f[u][j]<v)
			{
				u=f[u][j];
				times+=pow(2,j);
			}
		}
		times++;
		printf("%lld\n",times);
	}
	return 0;
}
