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<bits/stdc++.h>
using namespace std;
int a[100000];

vector<int> nxt(vector<int> x,vector<int> y){
	vector<int> res(x.size());
	for(int i=0;i<x.size();i++)
		res[i]=x[y[i]];
	return res;
}

vector<int> power(vector<int> x,long long int k){
	vector<int> res(x.size());
	for(int i=0;i<res.size();i++)
		res[i]=i;
	while(k){
		if(k&1)	res=nxt(res,x);
		x=nxt(x,x);
		k>>=1;
	}
	return res;
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin>>n;
	for(int i=0;i<n;i++)
		cin>>a[i];
	vector<int> x(n-1);
	for(int i=0;i<n-1;i++)
		x[i]=i;
	int m;
	long long int k;
	cin>>m>>k;
	for(int i=1,j;i<=m;i++){
		cin>>j;
		swap(x[j-2],x[j-1]);
	}
	x=power(x,k);
	long long int cur=a[0];
	cout<<cur<<'\n';
	for(int i=0;i<n-1;i++){
		cur+=a[x[i]+1]-a[x[i]];
		cout<<cur<<'\n';
	}
}