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
#pragma GCC optimize(3)
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int mod=998244353;
const int N=1e5+5;
int n,m,k,inv,a[N],b[N],cf[N],c[N],xu[N],f[N][65];
int xjh(int x,int y)
{
	for(int i=62;i>=0;i--)
		if(y&((long long)1<<i))x=f[x][i];
	return x;
}
signed main()
{
	scanf("%lld",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%lld",&a[i]);
		cf[i]=a[i]-a[i-1];
	}
	for(int i=1;i<=n;i++)a[i]=i;
	scanf("%lld%lld",&m,&k);
	for(int i=1;i<=m;i++)
	{
		scanf("%lld",&b[i]);
		swap(a[b[i]],a[b[i]+1]);
	}
	for(int i=1;i<=n;i++)f[i][0]=a[i];
	for(int i=1;i<=62;i++)
		for(int j=1;j<=n;j++)
			f[j][i]=f[f[j][i-1]][i-1];
	for(int i=1;i<=n;i++)xu[i]=cf[xjh(i,k)];
	double jia=0.0;
	for(int i=1;i<=n;i++)
	{
		jia=jia+(double)xu[i];
		printf("%.1lf\n",jia);
	}
	return 0;
}