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<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<iostream>
#define N 110000
#define LL long long
using namespace std;
int a[N],n,m,p[N],g[N],d[N],t[N];
LL k;
void qmod(LL b)
{
	for(int i=1;i<=n;i++) g[i]=i;
	while(b)
	{
		if(b%2)
		{
			for(int i=1;i<=n;i++) t[i]=g[p[i]];
			for(int i=1;i<=n;i++) g[i]=t[i];
		}
		b/=2;
		for(int i=1;i<=n;i++) t[i]=p[p[i]];
		for(int i=1;i<=n;i++) p[i]=t[i];
	}
	for(int i=1;i<=n;i++) t[i]=d[g[i]];
	for(int i=1;i<=n;i++) d[i]=t[i];
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) {scanf("%d",&a[i]);d[i]=a[i]-a[i-1];}
	for(int i=1;i<=n;i++) p[i]=i;
	scanf("%d%lld",&m,&k);
	for(int i=1;i<=m;i++)
	{
		int x;scanf("%d",&x);
		swap(p[x],p[x+1]);
	}
	qmod(k);
	LL now=0;
	for(int i=1;i<=n;i++)
	{
		now+=d[i];
		printf("%lld\n",now);
	}
	return 0;
}