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,f[100005],a[100005],L,s;
int q,p[100005][50];
int power(int a,int k){
	if(k == 0) return 1;
	if(k == 1) return a;
	int t = power(a,k >> 1);
	if(k % 2) return t * t * a;
	else return t * t;
}
int main(){
	scanf("%d",&n);
	for(int i = 1; i <= n; i++)scanf("%d",&a[i]);
	scanf("%d",&L);
	for(int i = 1,step = 1; i <= n; i++){
		while(a[i] - a[step] > L)step++;
		p[i][0] = step;
	}
	for(int i = 1,j = 2; j <= n; i++,j *= 2,s++){
		for(int k = 1; k <= n; k++){
			p[k][i] = p[p[k][i - 1]][i - 1];
		}
	}
	scanf("%d",&q);
	int x,y,j,k,ans;
	for(int i = 1; i <= q; i++){
		scanf("%d%d",&x,&y);
		if(x > y) swap(x,y);
		ans = 1;
		k = y;
		for(int j = s; j >= 0; j--){
			if(p[k][j] > x){
				k = p[k][j];
				ans += power(2,j);
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}