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>
using namespace std;
#define N 100050
double a[N],c[N],d[N];
int p[N],b[N],n,m,i,j,top,sta[N];
long long k;
bool vis[N];
int main()
{
	scanf("%d",&n);
	for (i=1;i<=n;i++) scanf("%lf",&a[i]);
	for (i=1;i<n;i++) p[i]=i;
	scanf("%d%lld",&m,&k);
	while (m--){
		int x;scanf("%d",&x);
		swap(p[x-1],p[x]);
	}
	for (i=1;i<n;i++) if (!vis[i]){
		top=0;
		sta[top]=i;vis[sta[top]]=true;
		while (p[sta[top]]!=i) sta[++top]=p[sta[top-1]],vis[sta[top]]=true;
		for (j=0;j<=top;j++) b[sta[j]]=sta[(j+k%(top+1))%(top+1)];
	}
	for (i=1;i<n;i++) c[i]=a[i+1]-a[i];
	for (i=1;i<n;i++) d[i]=c[b[i]];
	for (i=2;i<=n;i++) a[i]=a[i-1]+d[i-1];
	for (i=1;i<=n;i++) printf("%.10f\n",a[i]);
	return 0;
}