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 s64;
#define rep(i,l,r) for(int i=l;i<=r;++i)
const int N=1e5+5;
int px[N],p[N],d0[N],d1[N];

bool vis[N];
int c[N],len;
void pow_p(int n,s64 k)
{
	rep(x,1,n)
	if(!vis[x])
	{
		len=0;
		for(int y=x;!vis[y];y=p[y])vis[c[len++]=y]=1;
		int now=k%len;
		rep(i,0,len-1)d1[c[i]]=d0[c[(i+now)%len]]; 	
	}
}

int main()
{
	//freopen("1.in","r",stdin);
	int n;
	cin>>n;
	rep(i,1,n)scanf("%d",px+i);
	rep(i,1,n-1)d0[i]=px[i+1]-px[i];
	int m;s64 k;
	cin>>m>>k;
	rep(i,1,n-1)p[i]=i;
	while(m--)
	{
		int a;
		scanf("%d",&a);
		swap(p[a-1],p[a]);
	}
	pow_p(n-1,k);
	s64 ans=px[1];
	printf("%lld\n",ans);
	rep(i,1,n-1)
	{
		ans+=d1[i];
		printf("%lld\n",ans);
	}
}