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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

int N,M;long long K;
int X[100005],D[100005];

int P[61][100005];
int ans[100005];

int main(){
	scanf("%d",&N);
	for(int i=1;i<=N;i++) scanf("%d",&X[i]),D[i]=X[i]-X[i-1],P[0][i]=i;
	scanf("%d%lld",&M,&K);
	while(M--){
		int x;scanf("%d",&x);
		swap(P[0][x],P[0][x+1]);
	}
	for(int i=1;i<=60;i++)
	for(int j=1;j<=N;j++)
		P[i][j]=P[i-1][P[i-1][j]];
	for(int j=1;j<=N;j++) ans[j]=j;
	for(int i=60;~i;i--)if((K>>i)&1)
	for(int j=1;j<=N;j++)
		ans[j]=P[i][ans[j]];
	for(int i=1;i<=N;i++) printf("%d\n",X[i]=X[i-1]+D[ans[i]]);
}