Problem: 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}.

Examples

Input

3
-1 0 2
1 1
2


Output

-1.0
1.0
2.0


Input

3
1 -1 1
2 2
2 2


Output

1.0
-1.0
1.0


Input

5
0 1 3 6 10
3 10
2 3 4


Output

0.0
3.0
7.0
8.0
10.0

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define int long long
typedef double db;
const int N=1e5+10;
int n,m,tot,st,d[N],x,id[N],nt[N],K,ans[N],q[N],cnt,vs[N],p[N];
signed main(){
	scanf("%lld",&n);
	for (int i=1;i<=n;i++)scanf("%lld",&p[i]);
	for (int i=2;i<=n;i++)d[i]=p[i]-p[i-1],id[i]=i;
	scanf("%lld%lld",&m,&K);
	for (int i=1;i<=m;i++){
		scanf("%lld",&x);
		swap(id[x],id[x+1]);
	}
	for (int i=2;i<=n;i++) nt[id[i]]=i;
	for (int i=2;i<=n;i++)
		if (!vs[i]){
			int cnt=0;
			for (int j=i;!vs[j];j=nt[j]){
				q[++cnt]=j;
				vs[j]=1;
			}
			x=K%cnt;int k=x%cnt+1;
			for (int j=1;j<=cnt;j++){
				ans[q[k]]=d[q[j]];
				k=k==cnt?1:(k+1);
			}
		}
	double pos=p[1];
	for (int i=1;i<=n;i++){
		pos+=ans[i];
		printf("%.10lf\n",(db)pos);
	}
	return 0;
}