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<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=100010;
int a[maxn],c[maxn],b[maxn],A[maxn],ans[maxn];
int AA[maxn];
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;++i){
		scanf("%d",&AA[i]);
/**/	if(i!=1)a[i]=AA[i]-AA[i-1];
		else a[i]=AA[i];
		c[i]=i;
	}
	int m;LL k;
	scanf("%d%lld",&m,&k);
	for(int i=1;i<=m;++i){
		int x;
		scanf("%d",&x);
		swap(c[x],c[x+1]);
	}
//	cout<<a[3]<<endl;
	for(int i=1;i<=n;++i){
		if(!b[i]){
			int cnt=0;
			for(int j=i;!b[j];j=c[j]){
	//			cout<<j<<endl;
				b[j]=1;
				A[++cnt]=j;
			}
			for(int j=1;j<=cnt;++j){
				ans[A[j]]=a[A[((LL)(j+k-1))%cnt+1]];
		//		cout<<ans[A[j]]<<endl;
//				cout<<A[(j+k-1)%cnt+1]<<endl;
			}
		}
	}
	LL sum=0;
	for(int i=1;i<=n;++i){
		//cout<<ans[i]<<endl;
		sum+=ans[i];
		printf("%lld.0\n",sum);
	}
	return 0;
}