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<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define maxn 300005
#define LL long long
using namespace std;

inline int rd(){
	int x=0,f=1;char c=getchar();
	while(c<'0' || c>'9') f=c=='-'?-1:1,c=getchar();
	while(c<='9' && c>='0') x=x*10+c-'0',c=getchar();
	return x*f;
}

int n,m,x[maxn],d[maxn],id[maxn],stk[maxn],top;
LL k,ans[maxn];
bool vis[maxn];

int main(){
	n=rd();
	for(int i=1;i<=n;i++) x[i]=rd(),d[i]=x[i]-x[i-1],id[i]=i;
	m=rd(); scanf("%lld",&k);
	for(int i=1,a;i<=m;i++) a=rd(),swap(id[a],id[a+1]);
	for(int i=1;i<=n;i++)
		if(!vis[i]){
			top=0;
			for(int j=i;!vis[j];j=id[j]) stk[++top]=j,vis[j]=1;
			for(int j=1;j<=top;j++) ans[stk[j]]=d[stk[(k+j-1)%top+1]];
		}
	for(int i=1;i<=n;i++) ans[i]+=ans[i-1],printf("%.1lf\n",(double)ans[i]);
	return 0;
}