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<vector>
#include<algorithm>
using namespace std;
#define MAXN 100010
int x[MAXN],fa[MAXN][20],step[MAXN];
vector<int>G[MAXN];
void dfs(int u,int p)
{
	step[u]=step[p]+1;
	for(int i=1,j=2;j<step[u];i++,j<<=1)
		fa[u][i]=fa[fa[u][i-1]][i-1];
	for(int i=0;i<G[u].size();i++)
		dfs(G[u][i],u);
}
int Solve(int a,int b)
{
	if(a>b) swap(a,b);
	int d=step[a]-step[b];
	int ret=0;
	for(int i=0;(1<<i)<=d;i++)
		if(d&(1<<i))
			a=fa[a][i];
	if(a>=b) return d;
	else if(a<b) return d+1;
}
int main()
{
	int n,L,q,a,b;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&x[i]);
	scanf("%d",&L);
	int l=1,r=1;
	while(l<n)
	{
		while(r<=n&&x[r]-x[l]<=L) r++;
		fa[l][0]=r-1;
		G[r-1].push_back(l);
		l++;
	}
	dfs(n,0);
	scanf("%d",&q);
	while(q--)
	{
		scanf("%d%d",&a,&b);
		printf("%d\n",Solve(a,b));
	}
}