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<iostream>
#include<cstdio>
#include<cstring>
namespace test
{
	typedef long long ll;
	const int N=100100;
	int n,m;
	template<typename T0,typename T1>void mul(T0 *a,T1 *b)
	{
		static T0 c[N];
		for(int i=1;i<=n;i++)
			c[i]=a[b[i]];
		for(int i=1;i<=n;i++)
			a[i]=c[i];
	}
	int a[N],c[N];
	ll f[N];
	ll ans,b;
	void initialize()
	{
		scanf("%d",&n);
		for(int i=1;i<=n;i++)
			scanf("%lld",f+i),a[i]=c[i]=i;
		ans=f[1];
		for(int i=1;i<n;i++)
			f[i]=f[i+1]-f[i];

		scanf("%d%lld",&m,&b);
		for(int i=1,x;i<=m;i++)
			scanf("%d",&x),std::swap(a[x-1],a[x]);
	}
	void qpow()
	{
		for(;b;b>>=1,mul(a,a))
			if(b&1)mul(c,a);
	}
	void solve()
	{
		initialize();
		qpow();
		mul(f,c);
		printf("%lld\n",ans);
		for(int i=1;i<n;i++)
			printf("%lld\n",ans+=f[i]);
	}
}
int main()
{
//	freopen("in","r",stdin);
//	freopen("out","w",stdout);
	test::solve();
	return 0;
}