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
  kt125*/

#include<stdio.h>
#include<stdlib.h>
#define SIZE 11
typedef struct{
    int size; //格納最大数
    int count; //データ数
    char data[SIZE]; //データの配列
    char *top; //topへのポインタ
} stack;
void initialize_stack(stack *stk); //スタックを初期化する関数
void print_stack(stack *stk); //スタックの中身をprintする関数
void push(stack *stk, char x); //スタックのpush操作をする関数
void pop(stack *stk); //スタックのpop操作wpする関数

stack stk;

int main(void){
    char key[12];
    char *ptr;
    scanf("%s", key); //入力を読み取る
    ptr = key; //keyの先頭アドレスを渡す
    initialize_stack(&stk);
    while(*ptr){ //ヌル文字に至るまで数字ならpush,'B'ならpop
        if( *ptr == '0' || *ptr == '1'){
            push(&stk, *ptr);
        }else{
            pop(&stk);
        }
        ptr++;
    }
    print_stack(&stk);
    return 0;
}

void initialize_stack(stack *stk){
    stk->count = 0;
    stk->size = SIZE;
    stk->top = &(stk->data[SIZE]); //データの一番後ろのアドレスを渡す
}

void pop(stack *stk){
    if(stk->count <= 0){ //データが空のときは何もしない
        return;
    }else{
        stk->top++;
        stk->count--;
    }
}

void push(stack *stk, char x){
    if(stk->count >= stk->size){ //エラー出力
        printf("Error:stack overflow. (x=%lf)\n", x);
        exit(1);
    }else{
        //topをずらして、そのアドレスに新たにxを追加
        stk->top--;
        *(stk->top) = x;
        stk->count++;
    }
}

void print_stack(stack *stk){
    char *p;
    p = &(stk->data[SIZE-1]);
    //stkには古いものが後ろに格納されているので後ろからprint
    for(int i=0; i<stk->count; i++){
        printf("%c", *p--);
    }
    printf("\n");
}