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<stdio.h>
#include<bitset>
#define For(i,a,b) for(register int i=a;i<=b;i++)
using namespace std;
const int maxn=1e5+1000;
int n,m,pos[maxn],x[maxn],cf[maxn],a,s[maxn];
long long ans[maxn],k;
bitset <maxn> vis;
inline void swap(int &a,int &b){int c=a;a=b;b=c;}
signed main(){
	scanf("%d",&n);
	For(i,1,n){
		scanf("%d",&x[i]);
		pos[i]=i;
		cf[i]=x[i]-x[i-1];
	}
	scanf("%d%lld",&m,&k);
	For(i,1,m){
		scanf("%d",&a);
		swap(pos[a],pos[a+1]);
	}
	For(i,1,n){
		if(vis[i]){
			continue ;
		}
		int cnt=0;
		for(int j=i;!vis[j];j=pos[j]) {
			s[++cnt]=j;
			vis[j]=1;
		}
		For(j,1,cnt) {
			ans[s[j]]=1ll*cf[s[(k+j-1)%cnt+1]];
		}
	}
	For(i,1,n){
		ans[i]+=ans[i-1];
		printf("%lld.0\n",ans[i]);
	}
	return 0;
}