There are N rabbits on a number line. The rabbits are conveniently numbered 1 through N. The coordinate of the initial position of rabbit i is x_i.

The rabbits will now take exercise on the number line, by performing sets described below. A set consists of M jumps. The j-th jump of a set is performed by rabbit a_j (2≤a_j≤N-1). For this jump, either rabbit a_j-1 or rabbit a_j+1 is chosen with equal probability (let the chosen rabbit be rabbit x), then rabbit a_j will jump to the symmetric point of its current position with respect to rabbit x.

The rabbits will perform K sets in succession. For each rabbit, find the expected value of the coordinate of its eventual position after K sets are performed.

Constraints

* 3≤N≤10^5
* x_i is an integer.
* |x_i|≤10^9
* 1≤M≤10^5
* 2≤a_j≤N-1
* 1≤K≤10^{18}

Input

The input is given from Standard Input in the following format:


N
x_1 x_2 ... x_N
M K
a_1 a_2 ... a_M


Output

Print N lines. The i-th line should contain the expected value of the coordinate of the eventual position of rabbit i after K sets are performed. The output is considered correct if the absolute or relative error is at most 10^{-9}.

Examples

Input

3
-1 0 2
1 1
2


Output

-1.0
1.0
2.0


Input

3
1 -1 1
2 2
2 2


Output

1.0
-1.0
1.0


Input

5
0 1 3 6 10
3 10
2 3 4


Output

0.0
3.0
7.0
8.0
10.0
#include<iostream>
#include<cstdio>
using namespace std;
int n,m,top,st[101010];
int id[101010],bok[101010];
long long k,ans[101010],a[101010];
void change(int x)
{
	top=0;
	st[++top]=x;bok[x]=1;
	while(!bok[id[x]]) bok[id[x]]=1,st[++top]=x=id[x];
	int k1=k%top;
	for(int i=1;i<=top-k1;i++)
		ans[st[i]]=a[st[i+k1]]-a[st[i+k1]-1];
	for(int i=top-k1+1;i<=top;i++)
		ans[st[i]]=a[st[i-top+k1]]-a[st[i-top+k1]-1];
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%lld",&a[i]),id[i]=i;
	scanf("%d%lld",&m,&k);
	for(int i=1,x;i<=m;i++)
	{
		scanf("%d",&x);
		swap(id[x+1],id[x]);
	}
	for(int i=1;i<=n;i++)
	if(!bok[i]) change(i);
	printf("%lld.0\n",a[1]);
	for(int i=2;i<=n;i++)
		printf("%lld.0\n",a[i]=a[i-1]+ans[i]);
	return 0;
}