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 <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define m_k make_pair
#define int long long
using namespace std;
const int N=1e5+100;
int n,m,k,a[N],b[N],f[N],c[N];
int fa[N][64],sum[N];
signed main()
{
	scanf("%lld",&n);
	for (int i=1;i<=n;i++) scanf("%lld",&f[i]);
	scanf("%lld%lld",&m,&k);
	for (int i=1;i<=m;i++) scanf("%lld",&a[i]);
	for (int i=1;i<=n;i++) b[i]=i;
	for (int i=1;i<=m;i++) swap(b[a[i]],b[a[i]+1]);
	for (int i=1;i<=n;i++) fa[i][0]=b[i];
	for (int j=1;j<=62;j++)
	{
		for (int i=1;i<=n;i++) fa[i][j]=fa[fa[i][j-1]][j-1];
	}
	for (int i=1;i<=n;i++)
	{
		int cnt=k,x=i;
		for (int j=62;j>=0;j--)
		{
			if (cnt>=(1ll<<j))
			{
				cnt-=(1ll<<j);
				x=fa[x][j];
			}
		}
		if (x==1) c[i]=f[x];
		else c[i]=f[x]-f[x-1];
	}
	for (int i=1;i<=n;i++) sum[i]=sum[i-1]+c[i];
	for (int i=1;i<=n;i++)
	{
		double ans=sum[i];
		printf("%.12lf\n",ans);
	}
}