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>
using namespace std;
#define LL long long

const int maxn=2e5+10;

int n,m,f[maxn],fk[maxn],dep[maxn];
LL k,x[maxn],y[maxn];

int main()
{
	int i,pos;
	scanf("%d",&n);
	for (i=1;i<=n;i++) scanf("%lld",&x[i]);
	for (i=n;i>=1;i--) x[i]-=x[i-1];
	for (i=1;i<=n;i++) f[i]=i;
	scanf("%d%lld",&m,&k);
	for (i=1;i<=m;i++)
	{
		scanf("%d",&pos);
		swap(f[pos],f[pos+1]);
	}
	for (i=1;i<=n;i++)
		if (!dep[i])
		{
			int u=i,v,move;
			while (1)
			{
				dep[f[u]]=dep[u]+1;
				u=f[u];
				if (u==i) break;
			}
			for (move=k%dep[i],v=i;move--;) v=f[v];
			for (fk[i]=v;u=f[u],v=f[v],u!=i;) fk[u]=v;
		}
	for (i=1;i<=n;i++) y[i]=x[fk[i]];
	for (i=1;i<=n;i++)
		y[i]+=y[i-1],printf("%lld\n",y[i]);
}