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 <iostream>
#include <cstdio>
#include <cmath>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
const int N=2e5+5;
long long a[N],n,m,k,vis[N],b[N],cnt,lop[N],ans[N],res[N];
int main(){
	scanf("%lld",&n);
	rep(i,1,n) scanf("%lld",&a[i]);
	per(i,n,1) a[i]=a[i]-a[i-1],b[i]=i;
	scanf("%lld%lld",&m,&k);
	rep(i,1,m){
		int R;
		scanf("%d",&R);
		swap(b[R],b[R+1]);
	}
	rep(i,1,n){
		if(vis[i]) continue;
		lop[cnt=1]=i;
		for(int pos=b[i];pos!=i;pos=b[pos]){
			vis[pos]=1;
			lop[++cnt]=pos;
		}
		rep(j,1,cnt) ans[lop[j]]=lop[(j+k-1)%cnt+1];
	}
	rep(i,1,n) res[i]=a[ans[i]];
	rep(i,1,n) printf("%lld.0\n",res[i]+=res[i-1]);
	return 0;
}
