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 <cstdlib>
#include <algorithm>
#include <vector>
#include <cstring>
#define SIZE 100005

using namespace std;
typedef long long int ll;

int A[SIZE];
int X[SIZE],T[SIZE];
bool use[SIZE];
int nd[SIZE];

int main()
{
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++) scanf("%d",&A[i]);
	for(int i=0;i<n-1;i++) X[i]=i;
	int m;
	ll K;
	scanf("%d %lld",&m,&K);
	for(int i=0;i<m;i++)
	{
		int a;
		scanf("%d",&a);a--;
		swap(X[a-1],X[a]);
	}
	for(int i=0;i<n-1;i++)
	{
		if(!use[i])
		{
			int now=i;
			int len=0;
			do
			{
				use[now]=true;
				nd[len++]=now;
				now=X[now];
			}while(now!=i);
			for(int j=0;j<len;j++)
			{
				int to=(int) (K%len);
				to=nd[(j+to)%len];
				T[nd[j]]=to;
			}
		}
	}
	ll sum=A[0];
	for(int i=0;i<n;i++)
	{
		printf("%lld\n",sum);
		int pos=T[i];
		sum+=A[pos+1]-A[pos];
	}
	return 0;
}