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}.

Examples

Input

3
-1 0 2
1 1
2


Output

-1.0
1.0
2.0


Input

3
1 -1 1
2 2
2 2


Output

1.0
-1.0
1.0


Input

5
0 1 3 6 10
3 10
2 3 4


Output

0.0
3.0
7.0
8.0
10.0
#include <bits/stdc++.h>
using namespace std;

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int n; cin >> n;
    vector<long long> x(n), d(n-1), e(n-1), to(n-1), sw(n-1);
    for (int i = 0; i < n; i++) {
        cin >> x[i];
        if (i) {
            d[i-1] = x[i] - x[i-1], to[i-1] = i-1;
        }
    }
    int m; long long k; cin >> m >> k;
    for (int i = 0; i < m; i++) {
        int a; cin >> a; --a;
        swap(to[a-1], to[a]);
    }
    while (k) {
        if (k & 1) {
            for (int i = 0; i < n-1; i++) e[i] = d[to[i]];
            swap(d, e);
        }
        for (int i = 0; i < n-1; i++) sw[i] = to[to[i]];
        swap(to, sw);
        k /= 2;
    }
    for (int i = 0; i < n; i++) {
        if (i) x[i] = x[i-1] + d[i-1];
        cout << x[i] << "\n";
    }
    return 0;
}