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)

//by OIerC
//Forca Barcelona!
#include<cstdio>
#include<algorithm>
#define rep(i, a, b) for (register int i=(a); i<=(b); ++i)
#define per(i, a, b) for (register int i=(a); i>=(b); --i)
using namespace std;
const int N=100005, P=998244353;
int a[N], b[N], c[N], n, ans;

inline int read()
{
 	int x=0,f=1;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}

void redouble(long long x)
{
	for (; x; x>>=1)
	{
		if (x&1)
		{
			rep(i, 1, n) c[i]=a[b[i]];
			rep(i, 1, n) a[i]=c[i];
		}
		rep(i, 1, n) c[i]=b[b[i]];
		rep(i, 1, n) b[i]=c[i];
	}
}

int main()
{
	n=read();
	rep(i, 1, n) a[i]=read(), b[i]=i;
	per(i, n, 1) a[i]-=a[i-1];
	int m=read(), k=read();
	rep(i, 1, m) 
	{
		int x=read();
		swap(b[x], b[x+1]);
	}
	redouble(k);
	rep(i, 1, n) ans=ans+a[i], printf("%d\n", ans);
	return 0;
}