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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define rrep(ri,n) for(int ri = (int)(n-1); ri >= 0; ri--)
#define rep2(i,x,n) for(int i = (int)(x); i < (int)(n); i++)
#define rrep2(ri,x,n) for(int ri = (int)(n-1); ri >= (int)(x); ri--)
#define repit(itr,x) for(auto itr = x.begin(); itr != x.end(); itr++)
#define rrepit(ritr,x) for(auto ritr = x.rbegin(); ritr != x.rend(); ritr++)
#define ALL(x) x.begin(), x.end()
using ll = long long;
using namespace std;

int main(){
  int n;
  cin >> n;
  if(n == 0){
    cout << 0 << endl;
    return 0;
  }
  stack<int> ans;
  while(n != 0){
    int na = abs(n);
    if(na % 2 == 0){
      ans.push(0);
    }else{
      ans.push(1);
      n--;
    }
    n /= -2;
  }
  while(!ans.empty()){
    cout << ans.top();
    ans.pop();
  }
  cout << endl;
  return 0;
}