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: Time Limit Exceeded)

#include <iostream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <algorithm>
#include <set>
#include <map>
#define vv(a, b, c, d) vector<vector<d> >(a, vector<d>(b, c))
#define vvi std::vector<std::vector<int> >
#define vvl std::vector<std::vector<ll> >
#define MODs 1000000007;
typedef long long int ll;
using namespace std;

std::vector<int> fn;
int solve(int a){
  if(a<0){
    for(int i=0;i<fn.size();i++){
      if(2*(fn[i]-1)/3 >= abs(a)){
        int rt = 2*(fn[i]-1)/3;
        for(int j=i;j>=0;j--) if(rt-abs(a)>=2*fn[j]) rt-=2*fn[j];
        return -rt;
      }
    }
  }else{
    for(int i=0;i<fn.size();i++){
      if((fn[i]-1)/3 >= abs(a)){
        int rt = (fn[i]-1)/3;
        for(int j=i;j>=0;j--) if(rt-abs(a)>=fn[j]) rt-=fn[j];
        return rt;
      }
    }
  }
  return 0;
}
int main(int argc, char const *argv[]) {
  ll N, reg=0, ans=0;
  std::cin >> N;
  std::vector<int> data;
  for(reg=1;reg<1e12;reg*=4) fn.push_back(reg);
  while(N!=0){
    reg = solve(N);
    N = N - reg;
    ans += abs(reg);
  }
  while(ans>0){
    data.push_back(ans%2==1?1:0);
    ans /= 2;
  }
  if(data.size()!=0) for(int i=data.size()-1;i>=0;i--) std::cout << data[i];
  else std::cout << 0;
  std::cout << '\n';
  return 0;
}