Problem: 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

Solution (in CPP):

Solution:
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=100005;

long long N,A[MAXN],M,K,ans[MAXN];
int P[MAXN],stk[MAXN],tp;
bool vis[MAXN];

int main()
{
	scanf("%lld",&N);
	for(int i=1;i<=N;i++)
	{
		scanf("%lld",&A[i]);
		P[i]=i;
	}
	scanf("%lld%lld",&M,&K);
	for(int i=N;i>=1;i--)
		A[i]=A[i]-A[i-1];
	for(int i=1;i<=M;i++)
	{
		int x;
		scanf("%d",&x);
		swap(P[x],P[x+1]);
	}
	for(int i=1;i<=N;i++)
		if(!vis[i])
		{
			int u=i;
			tp=0;
			while(!vis[u])
			{
				stk[++tp]=u;
				vis[u]=true;
				u=P[u];
			}
			for(int j=1;j<=tp;j++)
				ans[stk[j]]=A[stk[(K+j-1)%tp+1]];
		}
	for(int i=1;i<=N;i++)
		ans[i]+=ans[i-1];
	for(int i=1;i<=N;i++)
		printf("%lld.0\n",ans[i]);
	
	return 0;
}//