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
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=1e5+5;
int n,m,i,j,x,p[N],a[N],t,q[N];
long long nk,v;
bool vis[N];
int read(){
	char c=getchar();int k=0,p=0;
	for (;c<48||c>57;c=getchar()) if (c=='-') p=1;
	for (;c>47&&c<58;c=getchar()) k=(k<<3)+(k<<1)+c-48;
	return p?-k:k;
}
int main(){
	for (n=read(),i=1;i<=n;i++) p[i]=read();
	m=read();scanf("%lld",&nk);
	for (i=1;i<n;i++) a[i]=i;
	for (;m--;){int x=read();swap(a[x-1],a[x]);}
	for (i=1;i<n;i++) if (!vis[i]){
		for (x=i,t=0;!vis[x];x=a[x]) vis[x]=1,q[t++]=x;
		for (j=0;j<t;j++) a[q[j]]=q[(j+nk)%t];
	}
	for (v=p[1],i=1;i<=n;i++)
		printf("%lld.0\n",v),v+=p[a[i]+1]-p[a[i]];
}