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:
/*
竟然转化转化成置换了^-^ 妙啊 ~
想起了哥哥之前说的“不同系统中的等效操作”，大致这个意思hhh
*/
#include <bits/stdc++.h>
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define per(i, x, y) for (int i = x; i >= y; i--)
using namespace std;

const int N = 1e5 + 10;
typedef long long ll;
ll n, m, K, pos[N], a[N], b[N], vis[N], p[N], ans[N];

int main() {
    cin >> n;
    rep(i, 1, n) scanf("%lld", &pos[i]), b[i] = i;
    per(i, n, 1) pos[i] -= pos[i - 1];
    cin >> m >> K;
    rep(i, 1, m) scanf("%lld", &a[i]), swap(b[a[i]], b[a[i] + 1]);
    rep(i, 1, n) if (!vis[i]) {
        int cnt = 0;
        int j = i;
        for (; !vis[j]; j = b[j]) {
            vis[j] = 1;
            p[++cnt] = j;
        }
        rep(j, 1, cnt) ans[p[j]] = pos[p[(j + K % cnt - 1) % cnt + 1]];
    }
    rep(i, 1, n) ans[i] += ans[i - 1], printf("%lld.0\n", ans[i]);
    return 0;
}