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<bits/stdc++.h>
#define L long long
using namespace std;
int n,m,x[100010],a[100010];
L p;
struct orz
{
	int a[100010];
	inline orz operator*(orz x)
	{
		orz p;
		int i;
		for(i=1;i<n;i++)
		  p.a[i]=a[x.a[i]];
		return p;
	}
}u[70],w;
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int i,j;
	L k;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	  {
	   scanf("%d",&x[i]);
	   w.a[i]=u[0].a[i]=i;
	  }
	scanf("%d%lld",&m,&p);
	for(i=1;i<=m;i++)
	  {
	   scanf("%d",&j);
	   swap(u[0].a[j],u[0].a[j-1]);
	  }
	for(i=1;i<60;i++)
	  u[i]=u[i-1]*u[i-1];
	for(i--;i>=0;i--)
	  if(p&(1ll<<i))
	    w=w*u[i];
	for(i=1;i<n;i++)
	  a[i]=x[w.a[i]+1]-x[w.a[i]];
	for(i=0,k=x[1];i<n;i++,k+=a[i])
	  printf("%lld\n",k);
	return 0;
}