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>

using namespace std;

using ll = long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template<typename Q_temp>
using smaller_queue = priority_queue <Q_temp, vector<Q_temp>, greater<Q_temp> >;

const int INF = (int) 1e9;
const ll LINF = (ll) 4e18;
const ll MOD = (ll) (1e9 + 7);
const double PI = acos(-1.0);
const int limit = 100010;

#define REP(i,m,n) for(ll i = m; i < (ll)(n); ++i)
#define rep(i,n) REP(i, 0, n)
#define MP make_pair
#define YES(n) cout << ((n) ? "YES" : "NO") << endl
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible") << endl
#define NP(v) next_permutation(v.begin(),v.end())
#define debug(x) cout << #x << ":" << x << endl;
#define debug2(x) for(auto a : x) cout << a << " "; cout << endl;

vector<pii> around = {MP(1, 0), MP(-1, 0), MP(0, 1), MP(0, -1)};

//------------------------------------------------------

ll mod(ll a, ll b) { //a mod b
    if (a >= 0) return a % b;
    else return (b - (-a) % b) % b;
}

ll pow2(ll n) { //(-2)^n
    ll res = 1;
    rep(i, n) res *= -2;
    return res;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n;
    cin >> n;
    if (n == 0) {
        cout << 0 << endl;
        return 0;
    }
    ll sum = 0;
    stack<ll> st;

    rep(i, 40) { //(-2)^i の位
        ll div = 1 << (i + 1);
        ll digit = (mod(mod(n, div) - mod(sum, div), div) == 0 ? 0 : 1);
        sum += digit * pow2(i);
        st.push(digit);
    }

    while (st.top() == 0) st.pop();
    while (!st.empty()) {
        cout << st.top();
        st.pop();
    }
    cout << endl;
    return 0;
}