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
#include<bits/stdc++.h>
using namespace std;
#define nn 100008
#define ll long long
ll n,m,pos[nn],dif[nn];
ll ck;
ll ct[nn];
ll res[nn],cx[nn];
ll c[nn],d[nn];
void mul(ll a[],ll b[])
{
	for(int i=1;i<n;i++) c[i]=i;
	for(int i=1;i<n;i++) c[i]=a[i];
	for(int i=1;i<n;i++) d[i]=c[b[i]];
	for(int i=1;i<n;i++) a[i]=d[i];
}
void qpow()
{
	while(ck)
	{
		if(ck&1) mul(res,cx);
		mul(cx,cx);ck=ck/2;
	}
}
int main()
{
	scanf("%lld",&n);for(int i=1;i<=n;i++) scanf("%lld",&pos[i]);
	for(int i=2;i<=n;i++) dif[i-1]=pos[i]-pos[i-1],cx[i-1]=res[i-1]=i-1;
	scanf("%lld%lld",&m,&ck);
	while(m--)
	{
		ll x;scanf("%lld",&x);
		swap(cx[x-1],cx[x]);
	}
	
	qpow();
	
	ll ct=pos[1];
	for(int i=1;i<=n;i++) printf("%lld\n",ct),ct+=dif[res[i]];
	return 0; 
}