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)

/* ex4-1
   kas056*/

#include <stdio.h>
#include <stdlib.h>

#define SIZE 15

typedef struct{
    int size;//格納最大数
    int count;//現在のデータ数
    char data[SIZE];//データ配列
    char *top;//topのポインタ
    char *bottom;//一番下のポインタ
}stack;

void initialize(stack *stk){
    stk -> count = 0;
    stk -> size = SIZE;
    stk -> top = &(stk->data[SIZE]);//配列の最終要素の一つ手前
    stk -> bottom = &(stk->data[SIZE]);//初期配列の要素の一つ手前...(*)
}

void push(stack *stk, char s){
    if(stk->count >= stk -> size){//格納最大数を超えたらエラー
        fprintf(stderr, "Error: Stack overflow. (s=%c)\n", s);
    }
    else{
        stk -> top--;//一つ上にあげる(=一つ手前に移動)
        *(stk->top) = s;
        stk->count++;
    }
}

void pop(stack *stk){
    //popが指示されたとき、
    if(stk->count > 0){//要素があれば取り出すようにする。
        stk->top++;//一つ下に移動(=一つ後に移動)
        stk->count--;
    }
}

void print_stack(stack *stk){
    char *c;
    int i;
    c = stk->bottom-1;//配列の一番下のポインタを指定する。
    
    for(i = stk->count-1; i>-1; i--){//スタック内の文字数分操作
        printf("%c", *c--);//出力したら一つ上にあげる。
    }
}


stack stk;
int main(void){
    char sentence[SIZE];
    int i;

    initialize(&stk);

    scanf("%s", sentence);

    for(i=0; i<SIZE; i++){
        if(sentence[i] == '0' || sentence[i] == '1'){
            push(&stk, sentence[i]);//B以外はpush
        }
        else if(sentence[i] == 'B'){
            pop(&stk);//Bがでたらpop
        }
        else{
            continue;
        }
    }

    print_stack(&stk);

    return 0;
    

}