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 <vector>
#include <algorithm>
#include <cmath>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <iomanip>
#include <utility>
#include <tuple>
#include <functional>
#include <bitset>
#include <cassert>
#include <time.h>
#define int long long
#define endl '\n'
#define INF 1000000000000000000
#define all(a) a.begin(),a.end()
#define rep(i, n) for (int i = 0; i < (n); i++)
#define fi first
#define se second
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<ll, P> PP;
template<class T, class S> bool chmax(T &a, const S &b) { if (a < b) { a = b; return 1; } return 0; }
template<class T, class S> bool chmin(T &a, const S &b) { if (a > b) { a = b; return 1; } return 0; }
ll gcd(ll n, ll m) { return (m ? gcd(m, n%m) : n); }
ll lcm(ll n, ll m) { return n / gcd(n, m)*m; }

vector<int> perm_comp(vector<int> a, vector<int> b) {
	int n = b.size();
	vector<int> res(n);
	rep(i, n) res[i] = a[b[i]];
	return res;
}

vector<int> perm_pow(vector<int> base, int k) {
	int n = base.size();
	vector<int> ans(n);
	rep(i, n) ans[i] = i;
	while (k) {
		if (k & 1) ans = perm_comp(ans, base);
		base = perm_comp(base, base);
		k >>= 1;
	}
	return ans;
}


signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(20);
	//modcalc();

	int n;
	cin >> n;
	vector<int> x(n);
	rep(i, n) cin >> x[i];
	int m, k;
	cin >> m >> k;
	vector<int> a(m);
	rep(i, m) cin >> a[i];
	vector<int> d(n - 1);
	rep(i, n - 1) d[i] = x[i + 1] - x[i];
	vector<int> ans(n - 1);
	rep(i, n - 1) ans[i] = i;
	rep(i, m) swap(ans[a[i] - 2], ans[a[i] - 1]);
	ans = perm_pow(ans, k);
	int res = x[0];
	cout << res << endl;
	rep(i, n - 1) {
		res += d[ans[i]];
		cout << res << endl;
	}
}