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
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=100005;
int n,x[N],t[N],m;
ll k,d[N];

typedef array<int,N> perm;
perm operator*(const perm &l,const perm &r)
{
	perm x;
	for(int i=1;i<=n;i++)x[i]=r[l[i]];
	return x;
}
perm power(perm a,ll x)
{
	perm ans;
	for(int i=1;i<=n;i++)ans[i]=i;
	for(;x;x>>=1,a=a*a)
		if(x&1)ans=ans*a;
	return ans;
}

int main()
{
	scanf("%d",&n);
	perm f;
	for(int i=1;i<=n;i++)scanf("%d",x+i),t[i]=x[i]-x[i-1],f[i]=i;
	scanf("%d%lld",&m,&k);
	for(int a;m--;)
	{
		scanf("%d",&a);
		swap(f[a],f[a+1]);
	}
	f=power(f,k);
	for(int i=1;i<=n;i++)d[i]=t[f[i]];
	for(int i=1;i<=n;i++)printf("%lld\n",d[i]+=d[i-1]);
}
