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<bits/stdc++.h>
using namespace std;
#define int long long
const int Mod=998244353;
int a[100010],b[100010],n,m,k,sum,tmp[100010];
void power(int k)
{
	while(k)
	{
		if(k&1)
		{
			for(int i=1;i<=n;i++)tmp[i]=a[b[i]];
			for(int i=1;i<=n;i++)a[i]=tmp[i];
		}
		for(int i=1;i<=n;i++)tmp[i]=b[b[i]];
		for(int i=1;i<=n;i++)b[i]=tmp[i];
		k>>=1;
	}
	return;
}
signed main()
{
//	freopen("qwq.in","r",stdin);
//	freopen("qwq.out","w",stdout);
	scanf("%lld",&n);
	for(int i=1;i<=n;i++)scanf("%lld",&a[i]),b[i]=i;
	for(int i=n;i>=1;i--)a[i]=a[i]-a[i-1];
	scanf("%lld%lld",&m,&k);
	for(int i=1;i<=m;i++)
	{
		int x;
		scanf("%lld",&x);
		swap(b[x],b[x+1]);
	}
	power(k);
	for(int i=1;i<=n;i++)sum=sum+a[i],printf("%lld.0\n",sum);
}