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
    kento_titech*/
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#define SIZE 11
typedef struct {//配列におけるスタックを構造体として定義
    int count;
    char data[SIZE];
    char *tail;//配列の要素の末尾部分のポインタ
    char *head;//先頭部分のポインタ
} stack;

void initialize_stack(stack *stk);
void print_stack(stack *stk);
void push(stack *stk, char x);
void pop(stack *stk);

stack stk;

int main(void){
    initialize_stack(&stk);
    char input[SIZE];
    int i;
    scanf("%s",input);
    for(i=0;i<strlen(input);i++){
        if(input[i]=='1'||input[i]=='0'){//'1','0'であればpush
            push(&stk,input[i]);
        }
        else//'B'のときpop
            pop(&stk);
    }
    print_stack(&stk);
    printf("\n");
    return 0;
}

void initialize_stack(stack *stk){//スタックの初期化
    stk->count = 0;
    stk->head=&(stk->data[0]);
    stk->tail=&(stk->data[0]);
}

void push(stack *stk,  char x){  //pushの定義  
    *(stk->tail) = x;
    stk->tail++;//末尾を一つ後ろに
    stk->count++;
}

void pop(stack *stk){//popの定義
    if(stk->count > 0){
        stk->tail--;//末尾を一つ前に
        stk->count--;
    }
}

void print_stack(stack *stk){//配列の中身を出力
    char *p;
    int i;    
    p = stk->head;
    for(i=0;i< stk->count;i++)//先頭からすべての要素を出力
        printf("%c",*p++);
}