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<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
const int maxn = 1e6+7;
typedef long long ll;


ll n,m,pos[maxn],k,ans[maxn];
ll a[maxn],b[maxn],st[maxn];
ll move[maxn],vis[maxn];



int main()
{
	scanf("%lld",&n);
	for(int i=1;i<=n;i++)
		scanf("%lld",&pos[i]);
	for(int i=1;i<=n;i++)
		b[i]=pos[i]-pos[i-1],a[i]=i;
	scanf("%lld%lld",&m,&k);
	for(int i=1;i<=m;i++)
	{
		ll x;
		scanf("%lld",&x);
		swap(a[x],a[x+1]);
	}
	for(int t=1;t<=n;t++)
	{
		if(vis[t])continue;
		int round = 0;
		for(int i=t;!vis[i];i=a[i])vis[i]=1,st[++round]=i;
		for(int i=1;i<=round;i++) ans[st[i]] = b[ st[(i+k-1)%round +1]];
	}
	for(int i=1;i<=n;i++)
	{
		ans[i]+=ans[i-1];
		printf("%.1lf\n",(double)ans[i]);
	}
	return 0;
}