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 n,l,q,u,v;
int a[100055];
int P[100055][20];
signed main(){
//	freopen(".inp","r",stdin);
//	freopen(".out","w",stdout);
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin>>n;
	for(int i=1;i<=n;i++) cin>>a[i];
	cin>>l;
	for(int i=1;i<=n;i++){
		int far=a[i]+l;
		int k=lower_bound(a+1,a+1+n,far)-a;
		if(a[k]>far||k==n+1) k--;
		P[i][0]=k;
	}
	for(int j=1;1<<j<=n;j++) for(int i=1;i<=n;i++) P[i][j]=P[P[i][j-1]][j-1];
	cin>>q;
	while(q--){
		cin>>u>>v;
		if(v<u) swap(u,v);
		int lg=log2(n),ans=0;
		for(int i=lg;i>=0;i--){
			if(P[u][i]<v){
				ans+=(1<<i);
				u=P[u][i];
			}
		}
		cout<<ans+1<<endl;
	}
}