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 N = 100000 + 10;
int n, m, u, x[N], nxt[N], vis[N], d[N];
long long k, a[N], p[N];
int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &x[i]);
        a[i] = x[i] - x[i - 1];
        nxt[i] = i;
    }
    scanf("%d%lld", &m, &k); 
    for (int i = 1; i <= m; i++) {
        scanf("%d", &u);
        swap(nxt[u], nxt[u + 1]);
    }
    for (int i = 1; i <= n; i++) {
        if (vis[i]) continue;
        int cnt = 0;
        for (int j = i; !vis[j]; j = nxt[j]) {
            vis[j] = 1;
            d[++cnt] = j;
        }
        for (int j = 1; j <= cnt; j++) p[d[j]] = a[d[(j + k - 1) % cnt + 1]];
    }
    for (int i = 1; i <= n; i++) {
        p[i] += p[i - 1];
        printf("%lld.0\n", p[i]);
    }
    return 0;
}