#include<bits/stdc++.h>
using namespace std;
long long n,x[111111],p[111111],y[111111],siz[111111],cir[111111],m,k,sum;
bool vis[111111];
int main(){
	scanf("%lld",&n);
	for(int i=1;i<=n;i++)scanf("%lld",&x[i]),p[i]=i;
	for(int i=1;i<=n;i++)y[i]=x[i]-x[i-1];
	scanf("%lld%lld",&m,&k);
	for(int i=1,a;i<=m;i++)scanf("%lld",&a),swap(p[a],p[a+1]);
	for(int i=1,c,pos,cnt=0;i<=n;i++)if(!vis[i]){
		++cnt,c=0,pos=i;
		while(!vis[pos])vis[pos]=1,cir[pos]=cnt,++c,pos=p[pos];
		siz[cnt]=c;
	}
	for(int i=1;i<=n;i++){
		int t=k%siz[cir[i]],pos=i;
		while(t--)pos=p[pos];
		printf("%lld.0\n",y[pos]+sum),sum+=y[pos];
	}
}
### Prompt

In CPP, your task is to solve 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;
long long n,x[111111],p[111111],y[111111],siz[111111],cir[111111],m,k,sum;
bool vis[111111];
int main(){
	scanf("%lld",&n);
	for(int i=1;i<=n;i++)scanf("%lld",&x[i]),p[i]=i;
	for(int i=1;i<=n;i++)y[i]=x[i]-x[i-1];
	scanf("%lld%lld",&m,&k);
	for(int i=1,a;i<=m;i++)scanf("%lld",&a),swap(p[a],p[a+1]);
	for(int i=1,c,pos,cnt=0;i<=n;i++)if(!vis[i]){
		++cnt,c=0,pos=i;
		while(!vis[pos])vis[pos]=1,cir[pos]=cnt,++c,pos=p[pos];
		siz[cnt]=c;
	}
	for(int i=1;i<=n;i++){
		int t=k%siz[cir[i]],pos=i;
		while(t--)pos=p[pos];
		printf("%lld.0\n",y[pos]+sum),sum+=y[pos];
	}
}
```