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 5.4.1) (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
#define LL long long 
using namespace std;
LL n,m,k;
LL a[101001],to[101001],tmp,b[101001],size,ans[101001],s[101001];
bool vis[101001];

int main()
{
	scanf("%lld",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%lld",&b[i]);
		to[i]=i;
	}
	for(int i=1;i<=n;i++)a[i]=b[i]-b[i-1];
	scanf("%lld%lld",&m,&k);
	for(int i=1;i<=m;i++)
	{
		scanf("%lld",&tmp);
		swap(to[tmp],to[tmp+1]);
	}
	for(int i=1;i<=n;i++)
	{
		if(vis[i])continue;
		memset(s,0,sizeof(s));
		size=0;
		for(int j=i;!vis[j];j=to[j])
		{
			vis[j]=true;
			size++;
			s[size]=j;
		}
		for(int j=1;j<=size;j++)
		{
			ans[s[j]]=a[s[(k+j-1)%size+1]];
		}
	}
	for(int i=1;i<=n;i++)
	{
		ans[i]+=ans[i-1];
		printf("%lld\n",ans[i]);
	}
}