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<algorithm>
#include<iostream>
#include<cstdio>
#define MN 100000
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
int n,m,a[MN+5],p[MN+5],vis[MN+5],q[MN+5],top,b[MN+5];
long long K;
int main()
{
	n=read();
	for(int i=1;i<=n;++i) p[i]=read();
	for(int i=1;i<n;++i) a[i]=i;
	m=read();cin>>K;
	for(int i=1;i<=m;++i)
	{
		int x=read();
		swap(a[x-1],a[x]);	
	}
	for(int i=1;i<n;++i) if(!vis[i])
	{
		vis[q[top=1]=i]=1;
		for(int x=a[i];!vis[x];x=a[x]) vis[q[++top]=x]=1;
		int add=top-K%top;
		for(int j=1;j<=top;++j) b[q[(j+add-1)%top+1]]=p[q[j]+1]-p[q[j]];
	}
	long long s=p[1];
	for(int i=1;i<=n;++i) printf("%lld\n",s),s+=b[i];
	return 0;
}