Problem:
Score : 300 points

Problem Statement

Given an integer N, find the base -2 representation of N.

Here, S is the base -2 representation of N when the following are all satisfied:

S is a string consisting of 0 and 1.

Unless S = 0, the initial character of S is 1.

Let S = S_k S_{k-1} ... S_0, then S_0 \times (-2)^0 + S_1 \times (-2)^1 + ... + S_k \times (-2)^k = N.

It can be proved that, for any integer M, the base -2 representation of M is uniquely determined.

Constraints

Every value in input is integer.

-10^9 \leq N \leq 10^9

Input

Input is given from Standard Input in the following format:

N

Output

Print the base -2 representation of N.

Sample Input 1

-9

Sample Output 1

1011

As (-2)^0 + (-2)^1 + (-2)^3 = 1 + (-2) + (-8) = -9, 1011 is the base -2 representation of -9.

Sample Input 2

123456789

Sample Output 2

11000101011001101110100010101

Sample Input 3

0

Sample Output 3

0

Solution in C++14 (Clang 3.8.0) (judge verdict: Accepted)

//#include <bits/stdc++.h>
// utility
#include <bitset>
#include <tuple>
#include <typeinfo>
// error handling
#include <cassert>
// string
#include <cctype>
#include <cstring>
#include <string>
// container
#include <array>
#include <deque>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#include <vector>
// algorithm
#include <algorithm>
// iterator
#include <iterator>
// math
#include <cmath>
#include <complex>
#include <numeric>
#include <random>
// i/o
#include <cstdio>
#include <iomanip>
#include <iostream>

using namespace std;
#ifdef _DEBUG
#define _GLIBCXX_DEBUG
#include "dump.hpp"
#else
#define dump(...)
#endif

#define int long long
typedef __int128_t Int;
#define DBG 1
#define rep(i, a, b) for (int i = (a); i < (b); i++)
#define rrep(i, a, b) for (int i = (b)-1; i >= (a); i--)
#define loop(n) rep(loop, (0), (n))
#define all(c) begin(c), end(c)
const int INF =
    sizeof(int) == sizeof(long long) ? 0x3f3f3f3f3f3f3f3fLL : 0x3f3f3f3f;
const int MOD = (int)(1e9) + 7;
const double PI = acos(-1);
const double EPS = 1e-9;
#define fi first
#define se second
using pii = pair<int, int>;
// template<class T> ostream &operator<<(ostream &os,T &t){dump(t);return os;}
template <typename T, typename S>
istream &operator>>(istream &is, pair<T, S> &p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T> void printvv(const vector<vector<T>> &v) {
  cerr << endl;
  rep(i, 0, v.size()) rep(j, 0, v[i].size()) {
    if (typeid(v[i][j]) == typeid(INF) and v[i][j] == INF) {
      cerr << "INF";
    } else
      cerr << v[i][j];
    cerr << (j == v[i].size() - 1 ? '\n' : ' ');
  }
  cerr << endl;
}

#ifndef _DEBUG
#define printvv(...)
#endif
void YES(bool f) { cout << (f ? "YES" : "NO") << endl; }
void Yes(bool f) { cout << (f ? "Yes" : "No") << endl; }
template <class T> bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T> bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(12);

  int N;
  cin >> N;
  if(N==0){
    cout<<0<<endl;
    return 0;
  }

  int x = 0, i = (N < 0);
  N = abs(N);
  while (x < N) {
    x |= (1ll << i);
    i += 2;
  }
  bitset<32> b(x ^ (x - N));
  stringstream ss;
  ss << b;
  cout << ss.str().substr(ss.str().find('1')) << endl;

  return 0;
}