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 p[100001][17],a[100001];
int A,B;

bool check(int x){
	int tmp=B,stp=0;
	while(x){
		if(x&1)	tmp=p[tmp][stp];
		stp++;
		x>>=1;
	}
	return tmp<=A;

}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n,L,k;
	cin>>n;
	for(int i=1;i<=n;i++)
		cin>>a[i];
	int tmp=n;
	cin>>L>>k;
	for(int i=n;i>=1;i--){
		while(tmp>0&&a[i]-a[tmp]<=L)
			tmp--;
		p[i][0]=tmp+1;
	}
	for(int j=1;j<=16;j++)
		for(int i=1;i<=n;i++)
			p[i][j]=p[p[i][j-1]][j-1];
	for(int i=1;i<=k;i++){
		cin>>A>>B;
		if(A>B)	swap(A,B);
		int l=0,r=100000;
		while(l<r){
			int mid=(l+r)/2;
			if(check(mid))
				r=mid;
			else
				l=mid+1;
		}
		cout<<l<<'\n';
	}
}
