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: Runtime Error)

// Why am I so dumb? :c
// chrono::system_clock::now().time_since_epoch().count()
#include<bits/stdc++.h>

#define pb push_back
#define mp make_pair

#define all(x) (x).begin(), (x).end()

#define fi first
#define se second

using namespace std;

typedef long long ll;   

const int MAXN = (int)1e5 + 5;

ll arr[MAXN];

int id[MAXN];

int n, m;

ll k;

void solve() {
    scanf("%d", &n);

    for (int i = 1; i <= n; ++i) {
        scanf("%lld", &arr[i]);
    }

    scanf("%d %lld", &m, &k);

    for (int i = 1; i <= m; ++i) {
        scanf("%d", &id[i]);
    }

    assert(k <= (ll)1e7 && m * k <= (ll)1e7);

    for (int step = 1; step <= k; ++step) {
        for (int i = 1; i <= m; ++i) {
            int p = id[i];
            arr[p] = arr[p - 1]  + arr[p + 1] - arr[p];
        }
    }

    for (int i = 1; i <= n; ++i) {
        printf("%lld\n", arr[i]);
    }
}

int main() {
    int tt = 1;

    while (tt--) {
        solve();
    }

    return 0;
}