Problem:
Score : 800 points

Problem Statement

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}.

Sample Input 1

3
-1 0 2
1 1
2

Sample Output 1

-1.0
1.0
2.0

Rabbit 2 will perform the jump.
If rabbit 1 is chosen, the coordinate of the destination will be -2.
If rabbit 3 is chosen, the coordinate of the destination will be 4.
Thus, the expected value of the coordinate of the eventual position of rabbit 2 is 0.5×(-2)+0.5×4=1.0.

Sample Input 2

3
1 -1 1
2 2
2 2

Sample Output 2

1.0
-1.0
1.0

x_i may not be distinct.

Sample Input 3

5
0 1 3 6 10
3 10
2 3 4

Sample Output 3

0.0
3.0
7.0
8.0
10.0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma region include
#include <iostream>
#include <iomanip>
#include <stdio.h>

#include <sstream>
#include <algorithm>
#include <cmath>
#include <complex>

#include <string>
#include <cstring>
#include <vector>
#include <tuple>
#include <bitset>

#include <queue>
#include <complex>
#include <set>
#include <map>
#include <stack>
#include <list>

#include <fstream>
#include <random>
//#include <time.h>
#include <ctime>
#pragma endregion //#include
/////////
#define REP(i, x, n) for(int i = x; i < n; ++i)
#define rep(i,n) REP(i,0,n)
/////////
#pragma region typedef
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
#pragma endregion //typedef
////定数
const int INF = (int)1e9;
const LL MOD = (LL)1e9+7;
const LL LINF = (LL)1e18;
const double PI = acos(-1.0);
const double EPS = 1e-9;
/////////
using namespace::std;

void solve(){
	int N;
	cin >> N;
	vector<LL> X(N);
	for(int i=0;i<N;++i){
		cin >> X[i];
	}
	vector<LL> D(N-1);//元
	for(int i=0;i<N-1;++i){
		D[i] = X[i+1]-X[i];
	}
	LL M,K;
	cin >> M >> K;
	vector< vector<int> > dp(64,vector<int>(N-1));
	for(int i=0;i<N-1;++i){
		dp[0][i] = i;
	}
	dp[1] = dp[0];
	
	vector<int> AA;
	for(int i=0;i<M;++i){
		int A;
		cin >> A;
		AA.push_back( A );
		swap(dp[1][A-2], dp[1][A-1]);
	}
	sort(AA.begin(), AA.end());

	for(int k=2;k<64;++k){//ダブリング
		//dp[k-1]からdp[k]を作る
		for(int i=0;i<N-1;++i){
			int pos1 = dp[k-1][i];//pos1->i
			int pos2 = dp[k-1][pos1];//pos2->pos1
			dp[k][i] = pos2;//pos2->i
		}
	}
	LL num = K;
	int pow = 1;
	vector<int> ans = dp[0];
	while(num){
		if(num&1){
			vector<int> temp = ans;
			for(int i=0;i<N-1;++i){
				int pos1 = dp[pow][ i ];//pos1->i
				temp[i] = ans[pos1];
			}
			ans = temp;
		}
		num >>=1;
		++pow;
	}
	LL now = X[0];
	cout << now << endl;
	for(int i=0;i<N-1;++i){
		now += D[ ans[i] ];
		cout << now << endl;
	}
}

#pragma region main
signed main(void){
	std::cin.tie(0);
	std::ios::sync_with_stdio(false);
	std::cout << std::fixed;//小数を10進数表示
	cout << setprecision(16);//小数点以下の桁数を指定//coutとcerrで別	
	
	solve();
}
#pragma endregion //main()