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<algorithm>
#include<cmath>
#include<vector>
#define SF scanf
#define PF printf
#define MAXN 100010
using namespace std;
typedef long long ll;
ll k;
int n,m,t;
int gt[MAXN];
ll a[MAXN],pre[MAXN],ans[MAXN];
vector<int> num;
bool vis[MAXN];
void got(int x){
	vis[x]=1;
	if(num.size()>=1&&x==num[0])
		return ;
	num.push_back(x);
	got(gt[x]);
}
int main(){
	SF("%d",&n);	
	for(int i=1;i<=n;i++)
		SF("%lld",&a[i]);
	for(int i=1;i<=n;i++)
		pre[i]=a[i]-a[i-1];
	SF("%d%lld",&m,&k);
	for(int i=1;i<=n;i++)
		gt[i]=i;
	for(int i=1;i<=m;i++){
		SF("%d",&t);	
		swap(gt[t],gt[t+1]);
	}
	for(int i=1;i<=n;i++){
		if(vis[i]!=0)
			continue;
		got(i);
		int len=num.size();
		ll k1=k%len;
		for(int j=0;j<len;j++)
			ans[num[j]]=pre[num[(j+k1)%len]];
		num.clear();
	}
	ll sum=0;
	for(int i=1;i<=n;i++){
		sum+=ans[i];
		PF("%lld.0\n",sum);
	}
}