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 <cstdio>
#include <algorithm>
#define ll long long
#define rep(i,j,k) for (i=j;i<=k;i++)
#define down(i,j,k) for (i=j;i>=k;i--)
using namespace std;
const int N=1e5+5,K=70;
int n,m,i,d[N],p[N],a[N];
int lim,j,trans[N][K];
ll sum,x[N];
int main()
{
//	freopen("rabbit.in","r",stdin);
//	freopen("rabbit.out","w",stdout);
	scanf("%d",&n);
	rep(i,1,n) scanf("%lld",&x[i]),d[i]=x[i]-x[i-1];
	scanf("%d%lld",&m,&sum);
	rep(i,1,m) scanf("%d",&a[i]);
	rep(i,1,n) trans[i][0]=p[i]=i;
	rep(i,1,m) swap(trans[a[i]][0],trans[a[i]+1][0]);
	for (lim=0;(1ll<<lim)<sum;lim++);
	rep(j,1,lim)
		rep(i,1,n) trans[i][j]=trans[ trans[i][j-1] ] [j-1];
	down(j,lim,0)
	{
		if ((1ll<<j)>sum) continue;
		rep(i,1,n) p[i]=trans[p[i]][j];
		sum-=(1ll<<j);
	}
	rep(i,1,n) x[i]=x[i-1]+d[p[i]];
	rep(i,1,n) printf("%lld\n",x[i]);
	return 0;
}