Problem:
Score : 200 points

Problem Statement

Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the 0 key, the 1 key and the backspace key.

To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string:

The 0 key: a letter 0 will be inserted to the right of the string.

The 1 key: a letter 1 will be inserted to the right of the string.

The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted.

Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter 0 stands for the 0 key, the letter 1 stands for the 1 key and the letter B stands for the backspace key. What string is displayed in the editor now?

Constraints

1 ≦ |s| ≦ 10 (|s| denotes the length of s)

s consists of the letters 0, 1 and B.

The correct answer is not an empty string.

Input

The input is given from Standard Input in the following format:

s

Output

Print the string displayed in the editor in the end.

Sample Input 1

01B0

Sample Output 1

00

Each time the key is pressed, the string in the editor will change as follows: 0, 01, 0, 00.

Sample Input 2

0BB1

Sample Output 2

1

Each time the key is pressed, the string in the editor will change as follows: 0, (empty), (empty), 1.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	public static String Y = "Yes";
	public static String N = "No";
	public static long MOD = (long) (Math.pow(10, 9) + 7);
	public static Scanner sc = new Scanner(System.in);

	public static void main(String[] args) {
		char[] s = nc();

		String ans = "";

		for (char c : s) {
			if (c == '0') {
				ans += "0";
			}
			if (c == '1') {
				ans += "1";
			}
			if (c == 'B') {
				if (ans.length() >= 1) {
					ans = ans.substring(0, ans.length() - 1);
				}
			}
		}

		out(ans);
	}

	/*
	 * 以下メソッド集
	 */
	static int ni() {
		return sc.nextInt();
	}

	static long nl() {
		return sc.nextLong();
	}

	static double nd() {
		return sc.nextDouble();
	}

	static String n() {
		return sc.next();
	}

	static char[] nc() {
		return sc.next().toCharArray();
	}

	static int kaijo(int n) {
		if (n == 0 || n == 1) {
			return 1;
		} else {
			return n * kaijo(n - 1);
		}
	}

	static int fib(int n) {
		return (n == 1 || n == 0) ? n : fib(n - 2) + fib(n - 1);
	}

	static long lcm(long m, long n) {

		return m * n / gcd(m, n);
	}

	static int lcm(int m, int n) {

		return m * n / gcd(m, n);
	}

	static long gcd(long a, long b) {
		return (b == 0) ? a : gcd(b, a % b);
	}

	static int gcd(int a, int b) {
		return (b == 0) ? a : gcd(b, a % b);
	}

	static void out(String info) {
		System.out.println(info);
	}

	static void out(int info) {
		System.out.println(info);
	}

	static void out(double info) {
		System.out.println(info);
	}

	static void out(long info) {
		System.out.println(info);
	}

	static void out(char info) {
		System.out.println(info);
	}

	static void out(Object obj) {
		System.out.println(obj.toString());
	}

	static void outn(String info) {
		System.out.print(info);
	}

	static void outn(int info) {
		System.out.print(info);
	}

	static void outn(double info) {
		System.out.print(info);
	}

	static void outn(long info) {
		System.out.print(info);
	}

	static void outn(char info) {
		System.out.print(info);
	}

	static double max(double d, double e) {
		return Math.max(d, e);
	}

	static long max(long a, long b) {
		return Math.max(a, b);
	}

	static double min(double d, double e) {
		return Math.min(d, e);
	}

	static long min(long a, long b) {
		return Math.min(a, b);
	}
}

class XY {
	int h;
	int c;

	XY(int h, int c) {
		this.h = h;
		this.c = c;
	}
}