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)

/*~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=
*$* WRITER:kakitamasziru/OxOmisosiru *$*
~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=*/
#ifdef LOCAL_JUDGE
#define _GLIBCXX_DEBUG
#endif
#include <iostream> // cout, endl, cin
#include <string> // string, to_string, stoi
#include <vector> // vector
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility> // pair, make_pair
#include <tuple> // tuple, make_tuple
#include <cstdint> // int64_t, int*_t
#include <limits>//setprecision
#include <map> // map
#include <queue> // queue, priority_queue
#include <set> // set,multiset
#include <stack> // stack
#include <deque> // deque
#include <math.h>//pow,,,
#include <cmath>//abs,,,
#include <bitset> // bitset
#include <numeric> //accumulate,,,
#define endl "\n";
using namespace std;

const long long INF = 4000000000000000001;
const int inf = 1001001001;
long long MOD = 1000000007;

void chmax(long long &a,long long b){if(a < b) a = b;}
void chmin(long long &a,long long b){if(a > b) a = b;}

//Solve N^M. This, mod_pow use Iterative Square Method.
long long mod_pow(long long N, long long M) {
    if (M == 0) return 1;
    long long res = mod_pow((N * N) % MOD, M / 2);
    //When end-of-a bit is 1, times simple N.
    if (M & 1) res = (res * N) % MOD;
    return res %= MOD;
}
long long gcd(long long a, long long b) {
    if (b == 0) return a; else return gcd(b, a % b);
}
long long lcm(long long a, long long b) {
    return a / gcd(a, b) * b ;
}



int main() {
    ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    
    long long N;cin >> N;
    if(N == 0){
        cout << 0 << endl;
        return 0;
    }
    string ans = "";
    long long div = 1;
    while(N){
        if(N % (2*abs(div)) != 0){
            ans += "1";
            N -= div;
        }
        else{
            ans += "0";
        }
        div *= -2;
    }
    reverse(ans.begin(),ans.end());
    cout << ans << endl;
}