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 X first
#define Y second
#define pb push_back
typedef double db;
typedef long long ll;
typedef pair<int,int> P;
const int MAXN=1e5+10;
ll res[MAXN],k,top;
int vis[MAXN],st[MAXN];
int n,m,x,d[MAXN],nxt[MAXN];

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) 
		scanf("%d",&d[i]),nxt[i]=i;
	for(int i=n;i>=1;i--) d[i]=d[i]-d[i-1];
	scanf("%d%lld",&m,&k);
	for(int i=1;i<=m;i++)
		scanf("%d",&x),swap(nxt[x],nxt[x+1]);
	
	for(int i=1;i<=n;i++)
	{
		if(vis[i]) continue;
		top=0;
		for(int j=i;!vis[j];j=nxt[j])
		{
			st[++top]=j;
			vis[j]=1;
		}
		for(int j=1;j<=top;j++)
			res[st[j]]=d[st[(j+k-1)%top+1]];
	}
	for(int i=1;i<=n;i++)
		res[i]+=res[i-1],printf("%lld\n",res[i]);
	return 0;
}