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;
#define maxn 100005
int n,l,q,x[maxn],fa[maxn][18],sum[maxn],summ;
inline void pre()
{
	register int i,j=2;
	for(i=1;i<=n;++i)
	{
		while(summ+sum[j]-sum[j-1]<=l&&j<=n) summ+=sum[j]-sum[j-1],++j;
		fa[i][0]=j-1;
		summ-=sum[i+1]-sum[i];
	}
	for(i=1;i<=17;++i)
		for(j=1;j<=n;++j)
			fa[j][i]=fa[fa[j][i-1]][i-1];
}
inline void solve(int a,int b)
{
	if(a>b) swap(a,b);
	int ans=0;
	for(register int i=17;i>=0;--i)
		if(fa[a][i]<b) a=fa[a][i],ans+=(1<<i);
	printf("%d\n",++ans);
}
int main()
{
	scanf("%d",&n);
	register int i;
	for(i=1;i<=n;++i)
		scanf("%d",&x[i]),sum[i]=sum[i-1]+x[i]-x[i-1];
	scanf("%d%d",&l,&q);
	pre();
	int a,b;
	for(i=1;i<=q;++i)
		scanf("%d%d",&a,&b),solve(a,b);
	return 0;
}