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 N 100005
using namespace std;
int n,m,b[N],fl[N];long double a[N];long long k;
int u[N];
void work(int a[],int b[])
{
	int ans[N];
	for (int i=1;i<n;i++)
		ans[i]=a[b[i]];
	for (int i=1;i<n;i++)
		a[i]=ans[i];
}
void work(int u[],long long k)
{
	int ans[N];
	for (int i=1;i<n;i++)
		ans[i]=i;
	for (;k;k>>=1)
	{
		if (k&1) work(ans,u);
		work(u,u);
	}
	for (int i=1;i<n;i++)
		u[i]=ans[i];
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
		scanf("%Lf",&a[i]);
	
	scanf("%d%lld",&m,&k);
	for (int i=1;i<=m;i++)
		scanf("%d",&b[i]);
	for (int i=1;i<n;i++)
		u[i]=i;
	for (int i=1;i<=m;i++)
		swap(u[b[i]-1],u[b[i]]);
	work(u,k);
	double now=a[1];
	for (int i=1;i<=n;i++)
	{
		printf("%.9f\n",now);
		now+=a[u[i]+1]-a[u[i]];
	}
}