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

Solution (in CPP):

Solution:
#include <cstdio>
#include <cstring>
#include <algorithm>
#define R register
using namespace std;
typedef long long LL;
const int MAXN=100100;
int d[MAXN],pmt[MAXN];

void permute(int a[],int b[],int n)
{
	int t[MAXN];
	for(R int i=1;i<=n;++i)
	  t[i]=a[i];
	for(R int i=1;i<=n;++i)
	  a[i]=t[b[i]];
}

void fpow(int a[],LL y,int n)
{
	while(y)
	{
		if(y&1)permute(d,a,n);
		permute(a,a,n),y>>=1;
	}
}


int main()
{
	int n;
	scanf("%d",&n);
	LL l;
	int t1,t2;
	scanf("%lld",&l);
	t1=l;
	for(R int i=2;i<=n;++i)
	{
		scanf("%d",&t2);
		d[i]=t2-t1;
		t1=t2;
	}
	for(R int i=1;i<=n;++i)
	  pmt[i]=i;
	int m;
	LL k;
	scanf("%d%lld",&m,&k);
	for(R int i=1;i<=m;++i)
	{
		scanf("%d",&t1);
		swap(pmt[t1],pmt[t1+1]);
	}
	fpow(pmt,k,n);
	for(R int i=1;i<=n;++i)
	{
		l+=d[i];
		printf("%lld\n",l);
	}
	return 0;
}