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>
const int N=1e5+5;
int x[N],n,m,i,j,b[N],c[N],z[N];
long long k,ans;
void solve(){
	if(k==0)return;
	int y=k&1;k>>=1;solve();
	for(memcpy(z+1,c+1,n-1<<2),i=1;i<n;++i)c[i]=z[c[i]];
	if(y)for(memcpy(z+1,c+1,n-1<<2),i=1;i<n;++i)c[i]=z[b[i]];
}
int main(){
	scanf("%d",&n);for(i=1;i<=n;++i)scanf("%d",x+i),b[i]=c[i]=i;for(i=n;i;--i)x[i]-=x[i-1];
	scanf("%d%lld",&m,&k);for(i=1;i<=m;++i)scanf("%d",&j),std::swap(b[j-1],b[j]);
	solve();printf("%lld\n",ans=x[1]);
	for(i=1;i<n;++i)printf("%lld\n",ans+=x[c[i]+1]);
	return 0;
}