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 JAVA):

Solution:
import java.io.*;
import java.util.Scanner;

class Main {
	private static int[] mulPerm(int[] a, int[] b, int len) {
		int[] c = new int[len + 1];

		for (int i = 1; i <= len; ++i) {
			c[i] = b[a[i]];
		}
		return c;
	}

	private static int[] powPerm(int[] a, int len, long k) {
		int[] p = new int[len + 1];
		for (int i = 1; i <= len; ++i) {
			p[i] = i;
		}

		while (k > 0) {
			if (k % 2 == 1) {
				p = mulPerm(p, a, len);
			}
			a = mulPerm(a, a, len);
			k /= 2;
		}
		return p;
	}

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);

		int n = in.nextInt();

		int[] a = new int[n + 1];
		int[] d = new int[n];

		for (int i = 1; i <= n; ++i) {
			a[i] = in.nextInt();
			if (i > 1) {
				d[i - 1] = a[i] - a[i - 1];
			}
		}

		int m = in.nextInt();
		long k = in.nextLong();

		int[] p = new int[n];
		
		for (int i = 1; i <= n - 1; ++i) {
			p[i] = i;
		}

		for (int i = 1; i <= m; ++i) {
			int index = in.nextInt();

			int tmp = p[index];
			p[index] = p[index - 1];
			p[index - 1] = tmp;
		}

		int[] q = powPerm(p, n - 1, k);

		long cur = a[1];
		for (int i = 1; i <= n - 1; ++i) {
			System.out.println(cur);
			cur += d[q[i]];
		}
		System.out.println(cur);
	}
}