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 <vector>
#include <algorithm>

using namespace std;
typedef long long ll;

int main(){
    ll n;
    cin >> n;

    ll s;
    if(n == 0){
        cout << 0 << endl;
        return 0;
    }else if(n < 0){
        for(s = 1; abs(n) > 1 << s; s += 2){}
        if(abs(n) == 1 << s){
            cout << 1;
            for(int i = 0; i < s; ++i){ cout << 0; }
            cout << endl;
            return 0;
        }
        n += 1 << s;
    }else{
        s = -1;
    }

    vector<int> bits;
    for(int i = 0; n > 0; ++i){
        if(n % 2){
            bits.push_back(1);
        }else{
            bits.push_back(0);
        }
        n /= 2;
    }

    vector<int> out(max((ll)bits.size() + 3, s + 1));
    if(s > 0){ out.at(s) = 1; }
    for(int i = 0; i < bits.size(); ++i){
        if(i % 2 == 0 && bits.at(i)){
            out.at(i) += 1;
        }else if(bits.at(i)){
            out.at(i) += 1;
            out.at(i + 1) += 1;
        }
        if(out.at(i) == 2){
            if(out.at(i + 1)){
                out.at(i) = 0;
                out.at(i + 1) -= 1;
            }else{
                out.at(i) = 0;
                out.at(i + 1) = 1;
                out.at(i + 2) = 1;
            }
        }
    }

    reverse(out.begin(), out.end());
    bool start_flg = false;
    for(int i = 0; i < out.size(); ++i){
        start_flg |= (out.at(i) != 0);
        if(start_flg){
            cout << out.at(i);
        }
    }
    cout << endl;
    return 0;
}