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<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
const int N=100005;
vector<int> vct[N];
int n,m,i,j,a[N],p[N],bl[N],cnt,id[N],s[N];
ll K,x[N],d[N],e[N];
void dfs(int x)
{
	if(bl[x])
		return;
	bl[x]=cnt;
	vct[cnt].push_back(x);
	id[x]=s[cnt]++;
	dfs(p[x]);
}
int main()
{
	scanf("%d",&n);
	for(i=1;i<=n;++i)
	{
		scanf("%lld",x+i);
		d[i]=x[i]-x[i-1];
		a[i]=i;
	}
	scanf("%d%lld",&m,&K);
	for(i=1;i<=m;++i)
	{
		scanf("%d",&j);
		swap(a[j],a[j+1]);
	}
	for(i=1;i<=n;++i)
		p[a[i]]=i;
	cnt=0;
	for(i=1;i<=n;++i)
		if(!bl[i])
		{
			++cnt;
			dfs(i);
		}
	for(i=1;i<=n;++i)
		e[vct[bl[i]][(id[i]+K)%s[bl[i]]]]=d[i];
	for(i=1;i<=n;++i)
	{
		e[i]+=e[i-1];
		printf("%lld.0\n",e[i]);
	}
	return 0;
}