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<stdio.h>
#include<algorithm>
int p2[30];
int next[100010][30]={0};
int x[100010];
void init(void){
	p2[0]=1;
	for (int i=1;i<=20;i++) p2[i]=p2[i-1]*2;
}
int main(void){
	init();
	int n;
	scanf("%d",&n);
	for (int i=1;i<=n;i++){
		scanf("%d",x+i);
	}
	int l;
	scanf("%d",&l);
	int p=1;
	for (int i=1;i<n;i++){
		for (;p<=n&&x[i]+l>=x[p];p++);
		next[i][0]=p-1;
	}
	for (int j=1;j<=20;j++){
		for (int i=1;i<=n;i++)next[i][j]=next[next[i][j-1]][j-1];
	}
	int q;
	scanf("%d",&q);
	while (q--){
		int a, b;
		scanf("%d%d",&a,&b);
		if (a>b) std::swap(a,b);
		int ans=0;
		while (a<b){
			int j;
			for (j=1;next[a][j]>0&&next[a][j]<=b;j++);
			a=next[a][j-1];
			ans+=p2[j-1];
		}
		printf("%d\n",ans);
	}
}