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 <bits/stdc++.h>
typedef long long ll;
using namespace std;
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)

const int MAX_N=1e5;
int N,M;
ll K;

typedef vector<int> vec;

vec mul(vec a, vec b){
	vec ret(N-1);
	REP(i,N-1){
		ret[i]=a[b[i]];
	}
	return ret;
}

vec pw(vec n,ll m){
	vec ret(N-1);
	REP(i,N-1){
		ret[i]=i;
	}
	if (!m){
		return ret;
	}
	ret=pw(n,m/2);
	ret=mul(ret,ret);
	if (m&1){
		ret=mul(ret,n);
	}
	return ret;
}

int main(){
	ll x0,x1,x;
	cin>>N>>x0;
	x=x0;
	vec dif(N-1);
	REP(i,N-1){
		cin>>x1;
		dif[i]=x1-x0;
		swap(x0,x1);
	}
	cin>>M>>K;
	vec p(N-1);
	REP(i,N-1){
		p[i]=i;
	}
	REP(i,M){
		int a;
		cin>>a;
		swap(p[a-2],p[a-1]);
	}
	p=pw(p,K);
	dif=mul(dif,p);
	cout<<x<<endl;
	x1=x;
	REP(i,N-1){
		x1+=dif[i];
		cout<<x1<<endl;
	}
	return 0;
}