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<algorithm>
using namespace std;
#define N 100005
#define LL long long
#define rep(i,j,k) for(i=j;i<=k;++i)
LL n,m,i,j,K;
LL x[N],d[N],to[N],tmp[N],sum[N];
void read(LL &p){
	p=0; char x=getchar(); bool fu=false;
	if(x=='-') fu=true;
	while(x<'0' || x>'9'){x=getchar(); if(x=='-') fu=true;}
	while(x>='0'&&x<='9'){p=p*10+x-'0'; x=getchar();}
	if(fu) p=-p;
}
void quick_mi(LL b){
	for(;b;b>>=1){
		if(b&1){
			rep(i,1,n) sum[i]=to[sum[i]];
		}
		rep(i,1,n) tmp[i]=to[i];
		rep(i,1,n) to[i]=tmp[to[i]];
	}
}
int main(){
	read(n);   
	rep(i,1,n){
		read(x[i]);
		d[i]=x[i]-x[i-1];
		sum[i]=tmp[i]=to[i]=i;
	}
	read(m); read(K);
	rep(i,1,m){
		read(j);
		swap(tmp[j],tmp[j+1]); 
	}
	rep(i,1,n) to[tmp[i]]=i;
	quick_mi(K);
	rep(i,1,n) x[sum[i]]=d[i];
	rep(i,1,n){
		x[i]+=x[i-1];
		printf("%lld\n",x[i]);
	}
	return 0;
}