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<iostream>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
#define ll long long
using namespace std;

const int N=100005;

int n,m,t,pos,cnt;
ll k;
double ans;
int x[N],a[N],p[N],f[N],b[N];
bool vis[N];

int main(){
	int i,j;
	scanf("%d",&n);
	for (i=1;i<=n;i++) scanf("%d",&x[i]);
	for (i=n;i>=1;i--) x[i]-=x[i-1];
	for (i=1;i<=n;i++) p[i]=i;
	scanf("%d%lld",&m,&k);
	for (i=1;i<=m;i++) scanf("%d",&a[i]);
	for (i=1;i<=m;i++) swap(p[a[i]],p[a[i]+1]);
	for (i=1;i<=n;i++) if (!vis[i]){
		pos=i; cnt=0;
		a[cnt]=i;
		while (p[pos]!=i){
			pos=p[pos];
			a[++cnt]=pos;
		}
		f[i]=a[k%(cnt+1)];
		for (j=1;j<=cnt;j++) f[a[j]]=p[f[a[j-1]]];
		for (j=0;j<=cnt;j++) vis[a[j]]=1;
	}
	for (i=1;i<=n;i++){
		ans+=(double)x[f[i]];
		printf("%.1f\n",ans);
	}
	return 0;
}