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<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define MAXN 100000
using namespace std;
typedef long long LL;
int n,m;
LL k;
LL x[MAXN+5];
int ord[MAXN+5];
int ord2[MAXN+5];
LL ans[MAXN+5],fin[MAXN+5];
bool vis[MAXN+5];
vector<int> ned;
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%lld",&x[i]),ord[i]=i;
		ans[i]=x[i]-x[i-1];
	}
	scanf("%d %lld",&m,&k);
	int p;
	for(int i=1;i<=m;i++)
	{
		scanf("%d",&p);
		swap(ord[p],ord[p+1]);
	}
	for(int i=1;i<=n;i++)
	{
		p=i;
		ned.clear();
		while(vis[p]==false)
		{
			vis[p]=true;
			ned.push_back(p);
			p=ord[p];
		}
		for(int j=0;j<(int)ned.size();j++)
			ord2[ned[j]]=ned[(1LL*j+1LL*k)%(LL)ned.size()];
	}
	for(int i=1;i<=n;i++)
		fin[i]=ans[ord2[i]];
	LL sum=0;
	for(int i=1;i<=n;i++)
	{
		sum+=fin[i];
		printf("%lld.0\n",sum);
	}
	return 0;
}