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 N=1e5+10;

bool vis[N];
int n,m,cnt,pos[N],a[N],node[N],to[N];
LL ans[N];
LL k;

void Init()
{
	scanf("%d",&n);
	for (int i=1;i<=n;++i) scanf("%d",&pos[i]);
	scanf("%d%lld",&m,&k);
	for (int i=1;i<=m;++i) scanf("%d",&a[i]);
}

void Dfs(int x)
{
	if (vis[x]) return ;
	vis[x]=1;
	node[cnt++]=x;
	Dfs(to[x]);
	
}

void Solve()
{
	for (int i=n;i>=2;--i) pos[i]-=pos[i-1];
	for (int i=1;i<=n;++i) to[i]=i;
	for (int i=1;i<=m;++i) swap(to[a[i]],to[a[i]+1]);
	
	for (int i=1;i<=n;++i)
		if (!vis[i])
		{
			cnt=0;
			Dfs(i);
			for (int j=0;j<cnt;++j)
				ans[node[j]]=pos[node[(j+k)%cnt]];
		}
	
	for (int i=2;i<=n;++i) ans[i]+=ans[i-1];
	
	for (int i=1;i<=n;++i) printf("%lld.0\n",ans[i]);
}

int main()
{
	Init();
	Solve();
	return 0;
}