//Love and Freedom.
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#define ll long long
#define db double
#define inf 20021225
#define N 100010
using namespace std;
int n,m,to[N],stk[N]; ll k,a[N],x[N],nw[N],ans[N]; bool vis[N];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)	scanf("%lld",&x[i]),nw[i]=x[i]-x[i-1],to[i]=i;
	scanf("%d%lld",&m,&k);
	for(int i=1;i<=m;i++)	scanf("%lld",&a[i]),swap(to[a[i]],to[a[i]+1]);
	for(int i=1;i<=n;i++)	if(!vis[i])
	{
		int cnt=0;
		for(int j=i;!vis[j];j=to[j])
			vis[j]=1,stk[++cnt]=j;
		for(int j=1;j<=cnt;j++)
			ans[stk[j]]=nw[stk[(j+k-1)%cnt+1]];
	}
	for(int i=1;i<=n;i++)
		ans[i]+=ans[i-1],printf("%lld\n",ans[i]);
	return 0;
}
### Prompt

Please formulate a Cpp solution to the following 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

### Response

```cpp
//Love and Freedom.
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#define ll long long
#define db double
#define inf 20021225
#define N 100010
using namespace std;
int n,m,to[N],stk[N]; ll k,a[N],x[N],nw[N],ans[N]; bool vis[N];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)	scanf("%lld",&x[i]),nw[i]=x[i]-x[i-1],to[i]=i;
	scanf("%d%lld",&m,&k);
	for(int i=1;i<=m;i++)	scanf("%lld",&a[i]),swap(to[a[i]],to[a[i]+1]);
	for(int i=1;i<=n;i++)	if(!vis[i])
	{
		int cnt=0;
		for(int j=i;!vis[j];j=to[j])
			vis[j]=1,stk[++cnt]=j;
		for(int j=1;j<=cnt;j++)
			ans[stk[j]]=nw[stk[(j+k-1)%cnt+1]];
	}
	for(int i=1;i<=n;i++)
		ans[i]+=ans[i-1],printf("%lld\n",ans[i]);
	return 0;
}
```