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: Accepted)

#include<cstdio>
int n,p[100010],m,x,r[100010],v[100010],beg[100010],cnt1,cnt2;
long long a[100010],k,ans[100010];
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i)scanf("%lld",a+i),a[i-1]=a[i]-a[i-1],p[i-1]=i-1;
	scanf("%d%lld",&m,&k);
	for(int i=1;i<=m;++i)scanf("%d",&x),p[x]^=p[x-1],p[x-1]^=p[x],p[x]^=p[x-1];
	for(int i=0;i<n;++i)if(!r[i]){
		v[beg[++cnt2]=++cnt1]=i;r[i]=cnt2;
		for(int j=p[i];j!=i;j=p[j])v[++cnt1]=j,r[j]=cnt2;
	}beg[cnt2+1]=n+1;
	for(int i=1;i<=cnt2;++i){
		int len=beg[i+1]-beg[i];
		for(int j=beg[i];j<beg[i+1];++j)ans[v[j]]=v[(j-beg[i]+k)%len+beg[i]];
	}
	ans[0]=a[ans[0]];for(int i=1;i<n;++i)ans[i]=a[ans[i]]+ans[i-1];
	for(int i=0;i<n;++i)printf("%lld.0\n",ans[i]);
	return 0;
}