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.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;
import java.util.concurrent.SynchronousQueue;

public class Main implements Runnable {
	public static void main(String[] args) {
		new Thread(null, new Main(), "", Runtime.getRuntime().maxMemory()).start();
		// new Main().run();
	}

	public void run() {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int[] x = new int[N];
		int[] d = new int[N];// d[i]: = | x[i] - x[i-1] |
		for (int i = 0; i < N; ++i) {
			x[i] = sc.nextInt();
			d[i] = x[i] - (i > 0 ? x[i - 1] : 0);
		}
		int M = sc.nextInt();
		long K = sc.nextLong();
		int[] a = new int[M];
		int[] comb = new int[N];// comb[i]:=i番目の数がcomb[i]に移動
		for (int i = 0; i < N; ++i) {
			comb[i] = i;
		}
		for (int i = 0; i < M; ++i) {
			a[i] = sc.nextInt();
			--a[i];
			// swap[ d[i] , d[i+1] ]
			{
				int tmp = comb[a[i]];
				comb[a[i]] = comb[a[i] + 1];
				comb[a[i] + 1] = tmp;
			}
		}

		int[] cur = new int[N];
		for (int i = 0; i < N; ++i) {
			cur[i] = i;
		}
		for (; K > 0; K >>= 1, comb = mul2(comb)) {
			if (K % 2 == 1) {
				cur = mul(cur, comb);
			}
		}
		long L = 0;
		for (int i = 0; i < N; ++i) {
			L += d[cur[i]];
			System.out.println(L);
		}
	}

	int[] mul(int[] ord1, int[] ord2) {
		int n = ord1.length;
		if (ord1.length != ord2.length) {
			throw new AssertionError();
		}
		int[] ret = new int[n];
		for (int i = 0; i < n; ++i) {
			ret[i] = ord1[ord2[i]];
		}
		return ret;
	}

	int[] mul2(int[] ord) {
		int n = ord.length;
		int[] ret = new int[n];
		for (int i = 0; i < n; ++i) {
			ret[i] = ord[ord[i]];
		}
		return ret;
	}

	static void tr(Object... objects) {
		System.out.println(Arrays.deepToString(objects));
	}
}