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
    takafumi_ueki */
#include <stdio.h>
#include <stdlib.h>
#define SIZE 11
typedef struct {
    int size;
    int count;
    char data[SIZE];
    char *top;
}stack;
void initialize_stack(stack *stk);//初期化
void print_stack(stack *stk);
void push(stack *stk, char key);
void pop(stack *stk);

int main(void){
    int i;
    char input[SIZE];
    stack string;
    //入力する
    scanf("%s",input);
    initialize_stack(&string);
    //0か1だったらその数字をプッシュし、Bが出たらpopする
    for(i=0;i<SIZE;i++){
        if(input[i] == '0'){
            push(&string,'0');
        }
        else if(input[i] == '1'){
            push(&string,'1');
        }
        else if(input[i]=='B'){
            pop(&string);
        }
    }
    //結果を出力する
    print_stack(&string);
    return 0;
}

void initialize_stack(stack *stk){
    stk->count = 0;
    stk->size = SIZE;
    stk->top = &(stk->data[SIZE]);
}

void push(stack *stk, char key){
    if(stk->count < stk->size){
        //トップをひとつ上にあげる
        stk->top--;
        //データを格納する
        *(stk->top) = key;
        //現在のデータ数をひとつ増やす
        stk->count++;
    }
}
void pop(stack *stk){
    if(stk->count > 0){
        //topをひとつ下に下げる
        stk->top++;
        //データ数をひとつ減らす
        stk->count--;
    }
}
void print_stack(stack *stk){
    char *ptr;
    int i;
    //一番上の位置
    ptr = stk->top;
    //古いデータからtopに向かってprintする
    while(i<stk->count-1){
        ptr++;
        i++;
    }
    for(int i=0;i<stk->count;i++){
        printf("%c", *ptr--);
    }
    printf("\n");
}