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 C (GCC 9.2.1) (judge verdict: Wrong Answer)

/* ex 9_1
   offNaria */

#include <stdio.h>
#include <stdlib.h>

//リストのノード（スタック内の各データ）の定義
typedef struct node
{
    char elem;
    struct node *next;
} node;

void push(node **first, char elem); //スタックにデータを追加
char pop(node **first);             //スタックからデータを取り出す

int main(void)
{
    char input[11], output[11];
    //最初のノードの指す先
    node *first = NULL;
    //スタックにノードを追加
    scanf("%s", input);
    for (int i = 0; i < 11; i++)
    {
        //'0'か'1'ならスタックに挿入
        if (input[i] == '0' || input[i] == '1')
        {
            push(&first, input[i]);
        }
        //'B'ならスタックの一番上のものを取り出す
        else if (input[i] == 'B')
        {
            pop(&first);
        }
        else
        {
            break;
        }
    }
    //スタックの逆順に出力
    for (int i = 0; i < 11; i++)
    {
        output[9 - i] = pop(&first);
    }
    for (int i = 0; i < 11; i++)
    {
        if (output[i] != 'x')
        {
            printf("%c", output[i]);
        }
    }
    printf("\n");
    return 0;
}

void push(node **first, char elem)
{
    //追加する分の領域を確保
    node *new = calloc(1, sizeof(node));
    if (new == NULL)
        exit(1);
    new->elem = elem;
    new->next = *first;
    *first = new;
}

char pop(node **first)
{
    char elem;
    node *tmp = *first;
    if (tmp != NULL)
    {
        elem = tmp->elem;
        *first = tmp->next;
        //取り出したノードのメモリを解放
        free(tmp);
        return elem;
    }
    //スタックの一番下またはスタックが空であれば目印記号'x'を返す
    else
    {
        return 'x';
    }
}