Problem:
Score : 800 points

Problem Statement

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}.

Sample Input 1

3
-1 0 2
1 1
2

Sample Output 1

-1.0
1.0
2.0

Rabbit 2 will perform the jump.
If rabbit 1 is chosen, the coordinate of the destination will be -2.
If rabbit 3 is chosen, the coordinate of the destination will be 4.
Thus, the expected value of the coordinate of the eventual position of rabbit 2 is 0.5×(-2)+0.5×4=1.0.

Sample Input 2

3
1 -1 1
2 2
2 2

Sample Output 2

1.0
-1.0
1.0

x_i may not be distinct.

Sample Input 3

5
0 1 3 6 10
3 10
2 3 4

Sample Output 3

0.0
3.0
7.0
8.0
10.0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define N 200010
#define int long long 
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,k,d[N],s[N],fin[N];
int now[N],cnt,rk[N],vis[N];
signed main()
{
	n=read();
	for(int i=1;i<=n;i++)
		d[i]=read();
	for(int i=2;i<=n;i++) s[i-1]=d[i]-d[i-1];
	m=read(); k=read();
	for(int i=1;i<n;i++) rk[i]=i;
	for(int i=1;i<=m;i++)
	{
		int x=read();
		swap(rk[x],rk[x-1]);
	} 
	//zhihuan
	for(int i=1;i<n;i++) if(!vis[i])
	{
		int x=rk[i]; 
		now[0]=i; 
		vis[i]=1;
		cnt=1;
		while(!vis[x])
		{
			vis[x]=1;
			now[cnt++]=x;
			x=rk[x];
		}
		for(int i=0;i<cnt;i++)
			fin[now[i]]=now[(i+k)%cnt];
	}
	
	printf("%lld\n",d[1]);
	for(int i=1;i<n;i++)
	{
		d[1]+=s[fin[i]];
		printf("%lld\n",d[1]);
	}
	return 0;
}