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<algorithm>
using namespace std;
inline int read()
{
	int x,f=1;char c;
	while((c=getchar())<'0'||c>'9')if(c=='-')f=0;
	for(x=c-'0';(c=getchar())>='0'&&c<='9';)x=x*10+c-'0';
	return f?x:-x;
}
#define MN 100000
long long x[MN+5],ans[MN+5],n,k;
struct data{int p[MN+5];}A,B;
data operator*(const data&x,const data&y)
{
	data res;
	for(int i=1;i<=n;++i)res.p[i]=y.p[x.p[i]];
	return res;
}
int main()
{
	int m,i;
	for(n=read(),i=1;i<=n;++i)x[i]=read(),A.p[i]=B.p[i]=i;
	for(i=n;i;--i)x[i]-=x[i-1];
	m=read();scanf("%lld",&k);
	while(m--)i=read(),swap(B.p[i],B.p[i+1]);
	for(;k;k>>=1,B=B*B)if(k&1)A=A*B;
	for(i=1;i<=n;++i)printf("%lld.0\n",ans[i]=ans[i-1]+x[A.p[i]]);
}