#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int n,m,x[N],a[N],b[N],c[N],p[N];long long kk;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
int main(){
	scanf("%d",&n);
	rep(i,1,n)scanf("%d",&x[i]);
	scanf("%d%lld",&m,&kk);
	rep(i,1,n-1)a[i]=b[i]=i;
	rep(i,1,m){int y;scanf("%d",&y);swap(a[y-1],a[y]);}
	for(;kk;kk>>=1){
		if(kk&1)
		rep(i,1,n-1)b[i]=a[b[i]];
		rep(i,1,n-1)c[i]=a[a[i]];
		rep(i,1,n-1)a[i]=c[i];
	}
	long long ans=x[1];
	rep(i,1,n){
		printf("%lld.0\n",ans);
		ans+=x[b[i]+1]-x[b[i]];
	}
}
### Prompt

Your challenge is to write 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
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int n,m,x[N],a[N],b[N],c[N],p[N];long long kk;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
int main(){
	scanf("%d",&n);
	rep(i,1,n)scanf("%d",&x[i]);
	scanf("%d%lld",&m,&kk);
	rep(i,1,n-1)a[i]=b[i]=i;
	rep(i,1,m){int y;scanf("%d",&y);swap(a[y-1],a[y]);}
	for(;kk;kk>>=1){
		if(kk&1)
		rep(i,1,n-1)b[i]=a[b[i]];
		rep(i,1,n-1)c[i]=a[a[i]];
		rep(i,1,n-1)a[i]=c[i];
	}
	long long ans=x[1];
	rep(i,1,n){
		printf("%lld.0\n",ans);
		ans+=x[b[i]+1]-x[b[i]];
	}
}
```