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<bits/stdc++.h>
using namespace std;

typedef long long LL;

const int MaxN=1e5+5;

int N,M;
LL K;
int X[MaxN],A[MaxN],D[MaxN];
int Base[MaxN],Ans[MaxN],Tmp[MaxN];

void Mul(int *a,int *b){
	int i;
	for(i=1;i<N;i++)
		Tmp[i]=b[a[i]];
	copy(Tmp+1,Tmp+N,a+1);
}

int main(){
	int i,j;
	LL now;
	scanf("%d",&N);
	for(i=1;i<=N;i++){
		scanf("%d",&X[i]);
		Base[i]=Ans[i]=Tmp[i]=i;
	}
	for(i=1;i<N;i++)
		D[i]=X[i+1]-X[i];
	scanf("%d%lld",&M,&K);
	for(i=1;i<=M;i++)
		scanf("%d",&A[i]);
	for(i=1;i<=M;i++)
		swap(Tmp[A[i]-1],Tmp[A[i]]);
	for(i=1;i<N;i++)
		Base[Tmp[i]]=i;
	for(;K;K>>=1){
		if(K&1)
			Mul(Ans,Base);
		Mul(Base,Base);
	}
	for(i=1;i<N;i++)
		Tmp[Ans[i]]=i;
	for(now=X[i=1];i<=N;i++){
		printf("%lld\n",now);
		now+=D[Tmp[i]];
	}
	return 0;
}