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.*;

public class Main {

	BufferedReader br;
	PrintWriter out;
	StringTokenizer st;
	boolean eof;

	void solve() throws IOException {
		int n = nextInt();
		double[] a = new double[n];
		for (int i = 0; i < n; i++) {
			a[i] = nextDouble();
		}
		
		int m = nextInt();
		long k = nextLong();
		
		int[] b = new int[m];
		
		for (int i = 0; i < m; i++) {
			b[i] = nextInt() - 1;
		}
		
		int[] perm = new int[n - 1];
		for (int i = 0; i < n - 1; i++) {
			perm[i] = i;
		}
		
		for (int pos : b) {
			pos--;
			int tmp = perm[pos];
			perm[pos] = perm[pos + 1];
			perm[pos + 1] = tmp;
		}
		
		int[] res = pow(perm, k);
		
		double[] deltas = new double[n - 1];
		for (int i = 0; i < n - 1; i++) {
			int idx = res[i];
			deltas[i] = a[idx + 1] - a[idx];
		}
		
		double tmp = a[0];
		out.println(tmp);
		for (double d : deltas) {
			tmp += d;
			out.println(tmp);
		}
	}
	
	int[] mult(int[] a, int[] b) {
		int[] c = new int[a.length];
		for (int i = 0; i < a.length; i++) {
			c[i] = b[a[i]]; // or???
//			c[i] = a[b[i]];
		}
		return c;
	}
	
	int[] pow(int[] p, long k) {
		int[] ret = new int[p.length];
		for (int i = 0; i < p.length; i++) {
			ret[i] = i;
		}
		
		for (; k > 0; k >>= 1) {
			if ((k & 1) == 1) {
				ret = mult(ret, p);
			}
			p = mult(p, p);
		}
		return ret;
	}

	Main() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		solve();
		out.close();
	}

	public static void main(String[] args) throws IOException {
		new Main();
	}

	String nextToken() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				eof = true;
				return null;
			}
		}
		return st.nextToken();
	}

	String nextString() {
		try {
			return br.readLine();
		} catch (IOException e) {
			eof = true;
			return null;
		}
	}

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
}