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<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define int long long
using namespace std;
inline int read(){
	int f=1,ans=0;char c=getchar();
	while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
	while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();}
	return f*ans;
}
const int MAXN=100001;
int m,k,ans,n,x[MAXN],d[MAXN],a[MAXN],fa[MAXN][61],p[MAXN];
signed main(){
//	freopen("2.in","r",stdin);
	n=read();
	for(int i=1;i<=n;i++) x[i]=read(),d[i]=i,p[i]=x[i]-x[i-1];
	m=read(),k=read();
	for(int i=1;i<=m;i++) a[i]=read(),swap(d[a[i]],d[a[i]+1]);
	for(int i=1;i<=n;i++) fa[i][0]=d[i];
	for(int j=1;j<=60;j++)
		for(int i=1;i<=n;i++) fa[i][j]=fa[fa[i][j-1]][j-1];
	for(int i=1;i<=n;i++){
		int res=i;
		int K=k;
		for(int j=60;j>=0;j--) if(K-(1ll<<j)>=0) K-=(1ll<<j),res=fa[res][j];
		ans+=p[res];
		printf("%lld.0\n",ans);
	}return 0;
}/*
3
1 -1 1
2 2
2 2
*/