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>
#define int long long
using namespace std;
const int maxn=1e5+10;
int c[maxn],a[maxn],to[maxn],ans[maxn],tmp[maxn],p[maxn],pos[maxn],K,now;
int n,m;
void solve(int y)
{
	for(;y;y>>=1)
	{
		if (y&1)
		{
			for (int i=1;i<=n;++i)p[i]=ans[to[i]];
			for (int i=1;i<=n;++i)ans[i]=p[i];
		}
		for (int i=1;i<=n;++i)p[i]=to[to[i]];
		for (int i=1;i<=n;++i)to[i]=p[i];
	}
	for(int i=1;i<=n;++i)p[i]=c[ans[i]];
	for(int i=1;i<=n;++i)c[i]=p[i];
}

signed main()
{
	cin>>n;
	for (int i=1;i<=n;++i)cin>>pos[i],c[i]=pos[i]-pos[i-1],to[i]=ans[i]=i;
	scanf("%d%lld",&m,&K);
	for(int i=1;i<=m;++i)cin>>a[i];
	for(int i=1;i<=m;++i)
		swap(to[a[i]],to[a[i]+1]);
	solve(K);
	for (int i=1;i<=n;++i)
		c[i]+=c[i-1],cout<<c[i]<<".0 "<<endl;;
}
/*
这个1e18 明示矩阵加速/倍增 
fi表示第i只兔子的期望位置。
扫操作，每次操作还要扫一下上一只和下一只，然后才能转移
这 
pos[i]=pos[i-1]+pos[i+1]-pos[i]

草啊谁tm想得到差分啊？？？？
 */