Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

const int Nmax = 1e5 + 5;
int a[Nmax], b[Nmax], c[Nmax], p[Nmax], x, n, i, m;
long long K;

void power(long long k)
{
    if(!k) return;

    int i;
    if(k&1)
    {
        for(i=1; i<n; ++i) c[i] = b[a[i]];
        for(i=1; i<n; ++i) b[i] = c[i];
    }

    for(i=1; i<n; ++i) c[i] = a[a[i]];
    for(i=1; i<n; ++i) a[i] = c[i];

    power(k/2);
}

int main()
{
  //  freopen("input", "r", stdin);
    cin.sync_with_stdio(false);

    cin >> n;
    for(i=1; i<=n; ++i) cin >> p[i], a[i] = i;

    cin >> m >> K;
    for(i=1; i<=m; ++i)
    {
        cin >> x;
        swap(a[x], a[x-1]);
    }

    p[0] = p[1];
    for(i=1; i<n; ++i) b[i] = i, p[i] = p[i+1] - p[i];
    power(K);

    long long s = p[0];
    for(i=1; i<=n; ++i)
    {
        cout << s << '\n';
        s += p[b[i]];
    }

    return 0;
}