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<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;

const int N=100005;

int n,T,l,s,t,r,mid;
int f[N][18],x[N];

void init(){
	int i,j;
	for (i=1;i<=n;i++)
		for (j=0;j<=17;j++) f[i][j]=n;
	for (i=1;i<n;i++) f[i][0]=lower_bound(x+1,x+1+n,x[i]+l+1)-x-1;
	for (j=1;j<=17;j++)
		for (i=1;i<=n;i++) f[i][j]=f[f[i][j-1]][j-1];
}

int go(int st,int d){
	int i;
	for (i=17;i>=0;i--)
		if (d>=(1<<i)){
			st=f[st][i];
			d-=(1<<i);
		}
	return st;
}

int main(){
	int i;
	scanf("%d",&n);
	for (i=1;i<=n;i++) scanf("%d",&x[i]);
	scanf("%d",&l);
	init();
	scanf("%d",&T);
	while (T--){
		scanf("%d%d",&s,&t); if (s>t) swap(s,t);
		l=1; r=n;
		while (l<r){
			mid=l+(r-l)/2;
			if (go(s,mid)>=t) r=mid;
			else l=mid+1;
		}
		printf("%d\n",r);
	}
	return 0;
}