#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>

using namespace std;
typedef long long LL;
const int maxn = 100005;

int x[maxn],p[maxn],c[maxn],w[maxn],n,M,d[maxn];LL K;

void mul(int *a,int *b)
{
	for (int i=1;i<n;i++) w[i]=b[a[i]];
	for (int i=1;i<n;i++) a[i]=w[i];
}
int main()
{
	#ifdef Amberframe
		freopen("agc006c.in","r",stdin);
		freopen("agc006c.out","w",stdout);
	#endif
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&x[i]);
	for (int i=1;i<n;i++) d[i]=x[i+1]-x[i];
	for (int i=1;i<n;i++) c[i]=p[i]=i;
	scanf("%d %lld",&M,&K);
	for (int u,i=1;i<=M;i++)
		scanf("%d",&u),swap(p[u],p[u-1]);
	while (K) {
		if (K&1) mul(c,p);
		mul(p,p);K>>=1;
	}
	LL cur=x[1];
	for (int i=1;i<=n;i++) printf("%lld\n",cur),cur+=d[c[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
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>

using namespace std;
typedef long long LL;
const int maxn = 100005;

int x[maxn],p[maxn],c[maxn],w[maxn],n,M,d[maxn];LL K;

void mul(int *a,int *b)
{
	for (int i=1;i<n;i++) w[i]=b[a[i]];
	for (int i=1;i<n;i++) a[i]=w[i];
}
int main()
{
	#ifdef Amberframe
		freopen("agc006c.in","r",stdin);
		freopen("agc006c.out","w",stdout);
	#endif
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&x[i]);
	for (int i=1;i<n;i++) d[i]=x[i+1]-x[i];
	for (int i=1;i<n;i++) c[i]=p[i]=i;
	scanf("%d %lld",&M,&K);
	for (int u,i=1;i<=M;i++)
		scanf("%d",&u),swap(p[u],p[u-1]);
	while (K) {
		if (K&1) mul(c,p);
		mul(p,p);K>>=1;
	}
	LL cur=x[1];
	for (int i=1;i<=n;i++) printf("%lld\n",cur),cur+=d[c[i]];
	return 0;
}
```