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<functional>
#include<algorithm>
#include<iterator>
#include<cstring>
#include<cstdio>
using namespace std;
int o[111111],r[111111],cr;
long long d[111111],x[111111];
bool viz[111111];
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
		scanf("%lld",&x[i]);
	for(int i=1;i<=n;++i)
		d[i]=x[i+1]-x[i];
	int m;
	long long k;
	scanf("%d%lld",&m,&k);
	for(int i=1;i<=n;++i)o[i]=i;
	for(int i=1;i<=m;++i)
	{
		int a;
		scanf("%d",&a);
		swap(o[a],o[a-1]);
	}
	for(int i=1;i<=n;++i)
		if(!viz[i])
		{
			cr=0;
			int j=i;
			do viz[r[cr++]=j]=1;while((j=o[j])!=i);
			for(j=0;j<cr;++j)
				o[r[j]]=r[(j+k)%cr];
		}
	for(int i=1;i<=n;++i)
		x[i]=d[o[i]];
	for(int i=n;i;--i)
		x[i]=x[i+1]-x[i];
	for(int i=1;i<=n;++i)
		printf("%lld.0\n",x[i]);
	return 0;
}