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<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
#define G() Cr=getchar()
int Xr;char Cr;
inline int rd(){   //读入优化
	Xr=0,G();
	while(Cr<'0'||Cr>'9')G();
	while(Cr>='0'&&Cr<='9')Xr=(Xr<<3)+(Xr<<1)+Cr-'0',G();
	return Xr;
}
#define MAX_N 100002
int n,va[MAX_N];
int L,q,l,r,ans;
int lg[20],dp[MAX_N][20]; 
int main(){
	n=rd();
	for(int i=1;i<=n;i++)va[i]=rd();
	L=rd();
	lg[0]=1;
	for(int i=1;i<=16;i++)lg[i]=lg[i-1]<<1; // 预处理 2^i
	
	int t=0;
	for(int i=1;i<=n;i++){
		while(va[t+1]-va[i]<=L&&t<n)t++;
		dp[i][0]=t;
	}
	
	for(int k=1;k<=16;k++){
		for(int i=1;i<=n;i++){
			dp[i][k]=dp[ dp[i][k-1] ][k-1];
		}
	}
	
	q=rd();
	while(q--){
		l=rd(),r=rd();
		if(l>r)swap(l,r);
		ans=0;
		for(int i=16;i>=0;i--)if(dp[l][i]<r)l=dp[l][i],ans+=lg[i];
		printf("%d\n",ans+1);
	}
}