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;

int P, N, M, x[100009], vv[100009], ans[100009], p[100009], ap[100009];
long long K;

int main ()
{
//freopen ("input", "r", stdin);
//freopen ("output", "w", stdout);

scanf ("%d", &N);
//M = 4, back (1);
//return 0;
for (int i=1; i<=N; i++)
    scanf ("%d", &x[i]), p[i] = i;
for (int i=N; i>=2; i--)
    x[i] -= x[i - 1];
scanf ("%d %lld", &M, &K);
for (int i=1; i<=M; i++)
{
    int j;
    scanf ("%d", &j);
    swap (p[j], p[j + 1]);
}
for (int i=1; i<=N; i++)
if (ap[i] == 0)
{
    int j = i, knt = 0;
    while (ap[j] == 0) ap[j] = 1, j = p[j], vv[knt ++] = j;
    int cat = K % knt;
    for (int p=0; p<knt; p++)
        ans[vv[p]] = vv[(p + cat) % knt];
}
long long curr = x[1];
printf ("%d.0\n", curr);
for (int i=2; i<=N; i++)
{
    curr += x[ans[i]];
    printf ("%lld.0\n", curr);
}

return 0;
}