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++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma gcc optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define PER(i, n) for (int i = (n-1); i >= 0; --i)
#define ALL(V) (V).begin(),(V).end()
#define SORT(V) sort(ALL(V)) //小さい方からソート
#define REV(V) reverse(ALL(V)) //リバース
#define RSORT(V) SORT(V);REV(V) //大きい方からソート
#define NEXP(V) next_permutation(ALL(V)) //順列
#define pb(n) push_back(n)
#define popb() pop_back()
#define endl '\n'
#define Endl '\n'
#define DUMP(x)  cout << #x << " = " << (x) << endl
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define yes(n) cout << ((n) ? "yes" : "no"  ) << endl
#define Yay(n) cout << ((n) ? "Yay!": ":(") << endl
#define VSUM(V) accumulate(ALL(V), 0)
#define MID(a,b,c) (a) < (b) && (b) < (c)
#define MIDe(a,b,c) (a) <= (b) && (b) <= (c)
#define IN(n) cin >> n
#define IN2(a,b) cin >> a >> b
#define IN3(a,b,c) cin >> a >> b >> c 
#define IN4(a,b,c,d) cin >> a >> b >> c >> d
#define VIN(V) for(int i = 0; i < (V).size(); i++) {cin >> (V).at(i);}
#define OUT(n) cout << n << endl
#define VOUT(V) REP(i, (V).size()) {cout << (V)[i] << endl;}
#define VOUT2(V) REP(i, (V).size()) {if((V).size()-1!=i){cout << (V)[i] << " ";}else{cout << (V)[i] << endl;}}

#define int long long
#define P pair<ll,ll>
#define Vi vector<ll>
#define VVi vector<vector<ll>>
#define Vd vector<double>
#define Vb vector<bool>
#define Vs vector<string>
#define Vc vector<char>
#define M map<ll,ll>
#define S set<ll>
#define PQ priority_queue<ll>
#define PQG priority_queue<ll,V,greater<ll>

using ll = long long;
using Graph = vector<vector<int>>;

const int MOD = 1000000007;
const int INF = 1061109567;
const double EPS = 1e-10;
const double PI  = acos(-1.0);
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};

class UnionFind {
public:
  vector <int> par;
  vector <int> siz;

  UnionFind(int sz_): par(sz_), siz(sz_, 1LL) {
    for (int i = 0; i < sz_; ++i) par[i] = i;
  }
  void init(ll sz_) {
    par.resize(sz_);
    siz.assign(sz_, 1LL);
    for (int i = 0; i < sz_; ++i) par[i] = i;
  }

  int root(int x) {
    while (par[x] != x) {
      x = par[x] = par[par[x]];
    }
    return x;
  }

  bool merge(int x, int y) {
    x = root(x);
    y = root(y);
    if (x == y) return false;
    if (siz[x] < siz[y]) swap(x, y);
    siz[x] += siz[y];
    par[y] = x;
    return true;
  }

  bool issame(int x, int y) {
    return root(x) == root(y);
  }

  int size(int x) {
    return siz[root(x)];
  }
};

int gcd(int a, int b) {
  return b != 0 ? gcd(b, a % b) : a;
}

int lcm(int a, int b) {
  return a * b / gcd(a, b);
}

// 文字を全て大文字にします
string toStrUp(string str) {
  char diff = 'A'-'a';
  REP(i,str.size()) str[i] += diff;
  return str;
}

// 文字をstring型で一文字取得します
string get1ch(string str, int key) {
  return str.substr(key,1);
}

// 素因数分解 (O(sqrt(n)))
map<int,int> prime_factor(int n) {
  map<int,int> ret;
  for(int i = 2; i * i <= n; i++) {
    while(n % i == 0) {
      ret[i]++;
      n /= i;
    }
  }
  if(n != 1) ret[n] = 1;
  return ret;
}

// 素数判定 (O(sqrt(n)))
bool is_prime(int x) {
  for(int i = 2; i * i <= x; i++) {
    if(x % i == 0) return false;
  }
  return true;
}

// 進数変換 (O(log n))
template<typename T>
vector<T> convert_base(T x, T b) {
  vector< T > ret;
  T t = 1, k = abs(b);
  while(x) {
    ret.emplace_back((x * t) % k);
    if(ret.back() < 0) ret.back() += k;
    x -= ret.back() * t;
    x /= k;
    t *= b / k;
  }
  if(ret.empty()) ret.emplace_back(0);
  reverse(begin(ret), end(ret));
  return ret;
}

template<class T> inline bool chmin(T& a, T b) {
  if(a > b) {
    a = b;
    return true;
  }
  return false;
}

template<class T> inline bool chmax(T& a, T b) {
  if(a < b){
    a = b;
    return true;
  }
  return false;
}

signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(15);

  // デフォルト変数定義
  int n=0,m=0,a=0,b=0,c=0,d=0,x=0,y=0,z=0;
  string s="",t="";
  //

  // ここから
  IN(s);
  string ans = "";
  REP(i,s.size()){
    if(s[i]=='0'){
      ans += "0";
    } else if(s[i]=='1'){
      ans += "1";
    } else {
      if(ans!=""){
        ans.erase(ans.begin()+ans.size()-1);
      }
    }
  }
  OUT(ans);
}