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<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=100005;
long long num[N],dif[N],pos[N];
long long mov[N],dfn[N],col[N];
vector<long long> cir[N];
long long n,m,k,cnt,las;
void DFS(long long nde,long long len)
{
	dfn[nde]=len,col[nde]=cnt;
	cir[cnt].push_back(nde);
	if(!col[pos[nde]])
		DFS(pos[nde],len+1);
}
int main ()
{
	scanf("%lld",&n);
	for(int i=1;i<=n;i++)
		scanf("%lld",&num[i]),pos[i]=i,dif[i-1]=num[i]-num[i-1];
	scanf("%lld%lld",&m,&k);
	for(int i=1;i<=m;i++)
		scanf("%lld",&mov[i]),swap(pos[mov[i]],pos[mov[i]-1]);
	for(int i=1;i<=n;i++)
		if(!dfn[i]) cnt++,DFS(i,0);
	las=num[1],printf("%lld\n",las);
	for(int i=1;i<n;i++)
		printf("%lld\n",las+=dif[cir[col[i]][(dfn[i]+k)%cir[col[i]].size()]]);
	return 0;
}