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 <cstdio>
#include <cstring>
#define MAXN 100010
#define LL long long

int n,m;
LL v[MAXN],ans[MAXN],K;
int a[MAXN];
int p[MAXN];
bool visit[MAXN];

void gao(int x){
	static int b[MAXN];
	int numb=0;
	b[numb++]=x;
	visit[x]=1;
	for(int i=a[x];i^x;i=a[i]){
		b[numb++]=i;
		visit[i]=1;
	}
	for(int i=0;i<numb;i++){
		int x=b[i];
		p[x]=b[(K+i)%numb];
	}
}

int main(){
#ifdef DEBUG
	freopen("C.in","r",stdin);
#endif
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%lld",v+i),a[i]=i;
	for(int i=n;i>=2;i--) v[i]-=v[i-1];
	scanf("%d%lld",&m,&K);
	while(m--){
		int t;
		scanf("%d",&t);
		a[t]^=a[t+1]^=a[t]^=a[t+1];
	}
	for(int i=1;i<=n;i++)
		if(!visit[i])
			gao(i);
	for(int i=1;i<=n;i++)
		ans[i]=v[p[i]];
	for(int i=2;i<=n;i++)
		ans[i]+=ans[i-1];
	for(int i=1;i<=n;i++) printf("%.1lf\n",(double)ans[i]);
}