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 <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
int main(){
	int n;	cin>>n;
	vector<int> x(n);
	vector<int> dif(n-1);
	for(int i=0;i<n;i++)	cin>>x[i];
	for(int i=0;i<n-1;i++)	dif[i]=x[i+1]-x[i];
	ll m,k;	cin>>m>>k;
	vector<int> pos(n-1);
	for(int i=0;i<n-1;i++)	pos[i]=i;
	for(int i=0;i<m;i++){
		int a;	cin>>a;	a--;
		swap(pos[a-1],pos[a]);
	}
	for(ll i=1;i<=k;i<<=1){
		if(k&i){
			vector<int> nextdif(n-1);
			for(int j=0;j<n-1;j++)	nextdif[j]=dif[pos[j]];
			dif=nextdif;
		}
		vector<int> nextpos(n-1);
		for(int j=0;j<n-1;j++)	nextpos[j]=pos[pos[j]];
		pos=nextpos;
	}
	ll ret=x[0];
	for(int i=0;i<n-1;i++){
		cout<<ret<<endl;
		ret+=dif[i];
	}
	cout<<ret<<endl;
	return 0;
}
