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;
typedef long long ll;
const int maxn=100005;
int n,L,Q,x[maxn],f[maxn][18],g[maxn][18];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",&x[i]);
	scanf("%d",&L);
	for(int i=1;i<=n;i++)
		f[i][0]=upper_bound(x+1,x+n+1,x[i]+L)-x-1;
	for(int j=1;j<=17;j++)
		for(int i=n;i;i--)
			f[i][j]=f[f[i][j-1]][j-1];
	for(int i=1;i<=n;i++)
		g[i][0]=lower_bound(x+1,x+n+1,x[i]-L)-x;
	for(int j=1;j<=17;j++)
		for(int i=1;i<=n;i++)
			g[i][j]=g[g[i][j-1]][j-1];
	int a,b;
	scanf("%d",&Q);
	while(Q--)
	{
		scanf("%d%d",&a,&b);
		int ans=0;
		if(a<b)
		{
			for(int i=17;i>=0;i--)
				if(f[a][i]<b) a=f[a][i],ans+=(1<<i);
		}
		else
		{
			for(int i=17;i>=0;i--)
				if(g[a][i]>b) a=g[a][i],ans+=(1<<i);
		}
		printf("%d\n",ans+1);
	}
	return 0;
}