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)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>

using namespace std;

const int MAXN = 100000 + 100;

int n, m;
long long k;
int init_x[MAXN];
int per[MAXN];
bool vis[MAXN];

int main()
{
	ios::sync_with_stdio(false);
	// freopen("1.in", "r", stdin);
	// freopen("1.out", "w", stdout);

	cin >> n;
	for(int i = 1; i <= n; i++)
		cin >> init_x[i];
	cin >> m >> k;
	for(int i = 1; i <= n - 1; i++)
		per[i] = i;
	for(int i = 1; i <= m; i++)
	{
		int x;
		cin >> x;
		swap(per[x - 1], per[x]);
	}
	for(int i = 1; i <= n - 1; i++)
		if(!vis[i])
		{
			vector<int> arr;
			int cur = i;
			do
			{
				vis[cur] = true;
				arr.push_back(cur);
				cur = per[cur];
			}
			while(cur != i);
			for(int i = 0; i < arr.size(); i++)
				per[arr[i]] = arr[(i + k) % arr.size()];
		}
	long long cur = init_x[1];
	cout << cur << endl;
	for(int i = 1; i <= n - 1; i++)
	{
		cur += init_x[per[i] + 1] - init_x[per[i]];
		cout << cur << endl;
	}

	return 0;
}