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;
const int N=100005;
int n,a[N],fa[22][N],L,q,x,y;
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++)scanf("%d",&a[i]);
	scanf("%d%d",&L,&q);
	for (int i=1;i<n;i++){
		int l=i+1,r=n;
		while (l<r){
			int mid=(l+r+1)/2;
			if (a[i]+L>=a[mid])l=mid;
			else r=mid-1;
		}
		fa[0][i]=l;
	} 
	fa[0][n]=n;
	for (int i=1;i<=19;i++)
		for (int j=1;j<=n;j++)fa[i][j]=fa[i-1][fa[i-1][j]];
	while (q--){
		scanf("%d%d",&x,&y);
		if (x==y){
			puts("0");
			continue;
		}
		if (x>y)swap(x,y);
		int ans=0;
		for (int i=19;i>=0;i--)
			if (fa[i][x]<y)x=fa[i][x],ans+=1<<i;
		printf("%d\n",ans+1);	
	}
}