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: Wrong Answer)

#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
const int N=100005;
LL K;
int A[N],B[N],P[N],d[N],t[N],i,j,k,n,m,x;
char c;

inline char getc()
{
	#define VV 10000000
	static char s[VV],*l=s,*r=s;
	if (l==r)
		l=s,r=s+fread(s,1,VV,stdin);
	return l==r?0:*l++;
}

LL read(){ int k=1; LL z=0; do c=getc(); while ((c<'0'||c>'9')&&c!='-'); if (c=='-') k=-1,c=getc(); while (c>='0'&&c<='9') z*=10,z+=c-'0',c=getc(); return k*z; }

void power(int *P,LL K)
{
	for (i=1;i<=n;++i) d[i]=i;
	for (;K;K>>=1)
	{
		if (K&1)
		{
			for (i=1;i<n;++i) t[i]=P[d[i]];
			for (i=1;i<n;++i) d[i]=t[i];
		}
		for (i=1;i<n;++i) t[i]=P[P[i]];
		for (i=1;i<n;++i) P[i]=t[i];
	}
}

int main()
{
	n=read();
	for (i=1;i<=n;++i) B[i]=read(),A[i-1]=B[i]-B[i-1],P[i]=i;
	m=read(),K=read();
	for (i=1;i<=m;++i) x=read(),swap(P[x-1],P[x]);
	power(P,K);
	for (i=1;i<=n;++i) B[i]=B[i-1]+A[d[i-1]],printf("%d.0\n",B[i]);
	return 0;
}