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 <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vll;
typedef pair<int, int> pii;
typedef vector<pii> vii;
#define sz(c) (int)(c).size()
#define ALL(c) (c).begin(), (c).end()

vi mult (const vi &p, const vi &q)
{
    const int n = sz(p);
    assert(n == sz(q));
    vi ans(n);
    for (int i = 0; i < n; i++)
        ans[i] = p[q[i]];
    return ans;
}

vi power (vi p, ll k)
{
    const int n = sz(p);
    vi ans(n);
    iota(ALL(ans), 0);

    while (k > 0)
    {
        if (k & 1)
            ans = mult(ans, p);
        p = mult(p, p);
        k >>= 1;
    }
    return ans;
}

void solve (int n)
{
    vi x(n);
    for (int i = 0; i < n; i++)
        cin >> x[i];

    int m;
    ll k;
    cin >> m >> k;

    vi a(m);
    for (int i = 0; i < m; i++)
    {
        cin >> a[i];
        --a[i];
    }

    vi df(n - 1);
    for (int i = 0; i < n - 1; i++)
        df[i] = x[i + 1] - x[i];

//    copy(ALL(df), ostream_iterator<int>(cerr, " "));
//    cerr << endl;

    vi perm(n - 1);
    iota(ALL(perm), 0);

    for (int i = 0; i < m; i++)
        swap(perm[a[i] - 1], perm[a[i]]);

    perm = power(perm, k);

//    copy(ALL(perm), ostream_iterator<int>(cerr, " "));
//    cerr << endl;

    vector<ll> ans(n);
    ans[0] = x[0];
    for (int i = 0; i < n - 1; i++)
        ans[i + 1] = ans[i] + df[perm[i]];

    for (int i = 0; i < n; i++)
        cout << ans[i] << '\n';
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    int n;
    while (cin >> n)
        solve(n);
}