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: Wrong Answer)

/* ex9_1
   UMAboogie */
#include <stdio.h>
#include <stdlib.h>
#define SIZE 10

typedef struct { //スタックを定義
  int size;
  int count;
  char data[SIZE];
  char *top;
} stack;

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

stack stk;

int main(void){

  int i = 0;
  char key[SIZE+1]; //一時的に格納する配列を宣言
  initialize_stack(&stk); //スタックを初期化
  scanf("%s", key); //入力されたデータをkeyに追加

  while(key[i] != '\0'){
    if(key[i] == 'B'){
      if(stk.count > 0){
        pop(&stk); //入力がBでかつcountが正であればpop
      }
    }else{
      push(&stk,key[i]); //入力が0or1のときそれをスタックに追加
    }
    i++;
  }
  print_stack(&stk); //結果を出力
  return 0;
}


void initialize_stack(stack *stk){ //スタックの初期化
  stk->count = 0; //要素数を0にする
  stk->size = SIZE; //スタックの大きさをSIZEにする
  stk->top = &(stk->data[SIZE]); //topを配列の末尾にする
}

void push(stack *stk, char x){ //データの追加
  if(stk->count >= stk->size){//オーバーフローする時
    exit(1);
  }else{
    stk->top--; //topを1つ手前にする
    *(stk->top) = x; //そこにデータを格納する
    stk->count++; //countを1加算する
  }
}

double pop(stack *stk){ //データの取り出し
  char x;
  if(stk->count <= 0){//アンダーフローする時
    exit(1);
  }else{
    x = *(stk->top); //topの位置からデータを取り出す
    stk->top++; //topを1つ下げる
    stk->count--; //countを1下げる
    return x;
  }
}

void print_stack(stack *stk){ //スタックの中身を古い順に出力
  for(int i=0;i<stk->count;i++){
    printf("%c",stk->data[SIZE-i-1]);
  }
  printf("¥n");
}