#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=1e6+1;
double s[N];
int n,m;ll k;
int a[N],p[N],u[N],tmp[N];
int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-f;ch=getchar();}
    while(isdigit(ch)){x=x*10+ch-48;ch=getchar();}
    return x*f;
}
int main(){
	n=read();
	for(int i=1;i<=n;i++) a[i]=read();
	m=read();scanf("%lld",&k);
	for(int i=1;i<=n;i++) p[i]=i,u[i]=i;
	for(int i=1;i<=m;i++){
		int x=read();
		swap(p[x],p[x+1]);
	}
	while(k){
		if(k&1){
			for(int i=1;i<=n;i++) tmp[i]=u[p[i]];
			for(int i=1;i<=n;i++) u[i]=tmp[i];
		}	
		for(int i=1;i<=n;i++) tmp[i]=p[p[i]];
		for(int i=1;i<=n;i++) p[i]=tmp[i];
		k>>=1;
	}
	for(int i=1;i<=n;i++) s[i]=a[u[i]]-a[u[i]-1]+s[i-1];
	for(int i=1;i<=n;i++) printf("%.1lf\n",s[i]);
	return 0;
}
### 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>
#define ll long long
using namespace std;
const int N=1e6+1;
double s[N];
int n,m;ll k;
int a[N],p[N],u[N],tmp[N];
int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-f;ch=getchar();}
    while(isdigit(ch)){x=x*10+ch-48;ch=getchar();}
    return x*f;
}
int main(){
	n=read();
	for(int i=1;i<=n;i++) a[i]=read();
	m=read();scanf("%lld",&k);
	for(int i=1;i<=n;i++) p[i]=i,u[i]=i;
	for(int i=1;i<=m;i++){
		int x=read();
		swap(p[x],p[x+1]);
	}
	while(k){
		if(k&1){
			for(int i=1;i<=n;i++) tmp[i]=u[p[i]];
			for(int i=1;i<=n;i++) u[i]=tmp[i];
		}	
		for(int i=1;i<=n;i++) tmp[i]=p[p[i]];
		for(int i=1;i<=n;i++) p[i]=tmp[i];
		k>>=1;
	}
	for(int i=1;i<=n;i++) s[i]=a[u[i]]-a[u[i]-1]+s[i-1];
	for(int i=1;i<=n;i++) printf("%.1lf\n",s[i]);
	return 0;
}
```