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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define repd(i,a,b) for (int i=(a);i<(b);i++)
#define rep(i,n) repd(i,0,n)
typedef long long ll;
using namespace std;
const int MOD = 1000000007;
const int INF = 1010000000;
const double EPS = 1e-10;
const array<pair<int,int>,4> fd{make_pair(1,0),make_pair(-1,0),make_pair(0,1),make_pair(0,-1)}; 

int main(){
  int n;cin>>n;
  string ans;
  while(n != 0){
    if(n > 0){
      int a = -(n/2);
      ans = (char)('0'+n-(-2*a)) + ans;
      n = a;    
    }else if (n < 0){
      int a = 0;
      while(true){
        if(n < (-2*a))a++;
        else break;
      }
      ans = (char)('0'+n-(-2*a)) + ans;n = a;
    }
  }
  if(n == 0){
    cout << 0 << endl;
  }else{
    cout << ans << endl;
  }
}