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;
#define N 1100000
#define ll long long
ll K;
int n,m,top;
int a[N],p[N],pos[N],vis[N],fin[N],st[N];
ll v[N],v1[N];
int main()
{
	//freopen("tt.in","r",stdin);
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%lld",&v[i]),p[i]=i;
	scanf("%d%lld",&m,&K);
	for(int i=1;i<=m;i++)
		scanf("%d",&a[i]);
	for(int i=1;i<=m;i++)
		swap(p[a[i]],p[a[i]+1]);
	for(int i=1;i<=n;i++)pos[p[i]]=i;
	for(int i=1,now;i<=n;i++)if(!vis[i])
	{
		st[0]=i;vis[i]=1;top=1;
		now=pos[i];
		while(now!=i)
		{
			st[top]=now;vis[now]=1;top++;
			now=pos[now];
		}
		int t=K%top;
		for(int j=0;j<top;j++)
			fin[st[(j+t)%top]]=st[j];
	}
	v1[1]=v[1];
	for(int i=2;i<=n;i++)v1[i]=v[i]-v[i-1];
	for(int i=1;i<=n;i++)v[i]=v1[fin[i]];
	v1[1]=v[1];
	for(int i=2;i<=n;i++)v1[i]=v[i]+v1[i-1];
	for(int i=1;i<=n;i++)
		printf("%lld.0\n",v1[i]);
	return 0;
}
