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>
using namespace std;
const long long MAXN=100005;
long long n,m,k,top;
long long a[MAXN],id[MAXN],st[MAXN],ans[MAXN];
bool vis[MAXN];
int main()
{
	scanf("%lld",&n);
	for(long long i=1;i<=n;i++)
	{
		scanf("%lld",&a[i]);
		id[i]=i;
	}
	for(long long i=n;i>=1;i--)
		a[i]-=a[i-1];
	scanf("%lld %lld",&m,&k);
	for(long long i=1;i<=m;i++)
	{
		long long x;
		scanf("%lld",&x);
		swap(id[x],id[x+1]);
	}
	for(long long i=1;i<=n;i++)
	{
		if(!vis[i])
		{
			top=0;
			for(long long j=i;!vis[j];j=id[j])
			{
				st[++top]=j;
				vis[j]=1;
			}
			for(long long j=1;j<=top;j++)
			{
				ans[st[j]]=a[st[(k+j-1)%top+1]];
			}
		}
	}
	for(long long i=1;i<=n;i++)
	{
		ans[i]+=ans[i-1];
		printf("%lld\n",ans[i]);
	}
}