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: Accepted)

/*  ex9_1
    Rikuta  */
#include <stdio.h>
//スタックに格納するデータの最大数
#define SIZE 10
//スタックの定義
typedef struct stack
{
    int size;
    int count;
    char data[SIZE];
    char *top;
} stack;
//スタックの初期化
void initialize_stack(stack *string);
//データの追加
void push(stack *string, char elem);
//データの取り出し
void pop(stack *string);
//stackの出力
void print_stack(stack *string);

int main(void)
{   
    //スタックを宣言
    stack string;
    //スタックを初期化
    initialize_stack(&string);
    while (1)
    {   
        //受け取る変数の宣言
        char temp_char;
        //入力
        scanf("%c", &temp_char);
        //入力が終了なら終わり
        if (temp_char=='\n')
        {
            break;
        }
        //Bが入力されたらスタックをpop
        if (temp_char=='B')
        {
            pop(&string);
        }
        //0,1ならそれをスタックにpush
        if (temp_char=='0'||temp_char=='1')        
        {
            push(&string, temp_char);
        }
    }
    //スタックを出力
    print_stack(&string);

    return 0;
}

//スタックの初期化
void initialize_stack(stack *string)
{   
    //スタックの要素数と最大の大きさとtopを初期化
    string->count = 0;
    string->size = SIZE;
    string->top = &(string->data[SIZE]);
}

//データの追加
void push(stack *string, char elem)
{   
    //スタックの長さが最大を超えていたら標準エラー出力
    if (string->count >= string->size)
    {
        fprintf(stderr, "Error: stack overflow. (x=%c)\n", elem);
        exit(1);
    }
    else
    {   
        //スタックのtopは一個戻り，戻ったtopに要素が入る．要素数が1増える．
        string->top--;
        *(string->top) = elem;
        string->count++;
    }
}

//データの取り出し
void pop(stack *string)
{   
    //データに何も入っていなかったら
    if (string->count > 0)
    {   
        //スタックのトップを奥へ，要素数が1減る
        string->top++;
        string->count--;
    }
}

void print_stack(stack *string)
{   
    //ポインタの宣言
    char *pointer;
    //ポインタはスタックの一番古い要素を指す
    pointer = string->top+string->count-1;
    for (int i = 0; i < string->count; i++)
    {   
        //ポインタがさす要素を新しくしながら出力していく．
        printf("%c", *(pointer--));
    }
    //改行
    printf("\n");
}