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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int i;
		long x[] = new long[N];
		for(i=0;i<N;i++){
			x[i] = sc.nextLong();
		}
		
		int M = sc.nextInt();
		long K = sc.nextLong();
		int a[] = new int[M];
		
		for(i=0;i<M;i++){
			a[i] = sc.nextInt();
		}
		
		long k;
		
		for(k=0;k<K;k++){
		
		//一回の体操
		for(i=0;i<M;i++){
			x[a[i]] = x[a[i]-1]+x[a[i]+1]-x[a[i]];
		}
		
		}

		for(i=0;i<N-1;i++){
			System.out.println(x[i]);
		}
	}

}