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;
typedef long long LL;
const int N=100005;
int n,m,x[N],a[N];
int b[N],vis[N],tmp[N],t;
int ans[N];
LL K;
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
		scanf("%d",&x[i]);
	scanf("%d%lld",&m,&K);
	for (int i=1;i<n;i++)
		b[i]=i;
	for (int i=1;i<=m;i++){
		scanf("%d",&a[i]);
		swap(b[a[i]-1],b[a[i]]);
	}
	memset(vis,0,sizeof vis);
	for (int i=1;i<n;i++){
		if (vis[i])
			continue;
		vis[i]=1;
		tmp[t=1]=i;
		for (int j=b[i];j!=i;j=b[j])
			vis[tmp[++t]=j]=1;
		int d=K%t;
		for (int j=1;j<=t;j++)
			ans[tmp[j]]=tmp[(j-1+d)%t+1];
	}
	LL tot=x[1];
	printf("%lld\n",tot);
	for (int i=1;i<n;i++)
		printf("%lld\n",tot+=x[ans[i]+1]-x[ans[i]]);
	return 0;
}