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
    L40S38  */
#include<stdio.h>
#include<stdlib.h>

typedef struct stack{
    char row[11];
    //入力される文字列は最大10文字+ヌル文字
    int letter_num;
    //文字列の長さ
}stack;

void push(stack *stack_p,char *letter){
    //stackの文字列の最後にletterを加える
    stack_p->row[stack_p->letter_num]=*letter;
    stack_p->letter_num++;
    return;
}

void pop(stack *stack_p){
    //stackの文字列の最後の文字を削除
    if(stack_p->letter_num!=0){
        stack_p->letter_num--;
        stack_p->row[stack_p->letter_num]='\0';
    }
    return;
}

int main(void){
    stack *Stack;
    Stack=(stack*)calloc(1,sizeof(stack));
    if(Stack==NULL){
        printf("ERROR");
        return -1;
    }
    char input[11];
    //input:入力された'0','1','B'のみでできた文字列

    int i=0;

    //入力された文字列をinputに格納
    scanf("%s",input);

    //input[i]ごとにStack中の文字列における処理を変更
    while(input[i]){
        if(input[i]=='B'){
            //バックスペースキーの時はpop操作
            pop(Stack);
        }
        else{
            //それ以外の時はpush操作
            push(Stack,&input[i]);
        }
        i++;
    }

    //結果の出力
    for(i=0;i<Stack->letter_num;i++){
        printf("%c",Stack->row[i]);
    }
    putchar('\n');
    free(Stack);
    return 0;

}