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<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<bitset>

using namespace std;

typedef long long LL;

int n,m;
LL k;
int x[110000];
int dx[110000];
int pp[110000];
LL ans[110000];
int a[110000],nn,id[110000];
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
		scanf("%d",&x[i]),dx[i-1]=x[i]-x[i-1],id[i]=i;
	scanf("%d %lld",&m,&k);
	for (int i=1;i<=m;i++)
	{
		int x;
		scanf("%d",&x);
		swap(id[x-1],id[x]);
	}
	for (int i=1;i<n;i++)
		if (!pp[i])
		{
			nn=0;
			int now=i;
			while (!pp[now]) a[nn++]=now,pp[now]=1,now=id[now];
			for (int j=0;j<nn;j++)
				ans[a[j]]=dx[a[((LL)j+k)%nn]];
		}
	ans[0]=dx[0];
	LL now=0;
	for (int i=1;i<=n;i++)
		printf("%lld.0\n",now+=ans[i-1]);
	return 0;
}