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: Accepted)

#include <bits/stdc++.h>

#define FAST ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define pf printf
#define sc scanf
#define sf(n) scanf("%d",&n)
#define sff(n1,n2) scanf("%d %d",&n1,&n2)
#define sfff(n1,n2,n3) scanf("%d %d %d",&n1,&n2,&n3)
#define sl(n) scanf("%lld",&n)
#define sll(n1,n2) scanf("%lld %lld",&n1,&n2)
#define slll(n1,n2,n3) scanf("%lld %lld %lld",&n1,&n2,&n3)
#define rep0(i,n) for(i=0;i<n;i++)
#define rep(i,n) for(i=1;i<=n;i++)
#define reps(i,a,n) for(i=a;i<=n;i++)
#define nl puts("");
#define pb push_back
#define MOD 1000000007
#define fi first
#define se second
#define N 100005
#define mem(ara,n) memset(ara,n,sizeof(ara))
#define memb(ara) memset(ara,false,sizeof(ara))
#define all(x) (x).begin(),(x).end()
#define sq(x) ((x)*(x))
#define sz(x) x.size()
#define pi pair<int,int>
#define pii pair<pair<int,int>,pair<int,int> >
#define line puts("-------");
#define db(x) cout<<#x<<" :: "<<x<<"\n";
#define dbb(x,y) cout<<#x<<" :: "<<x<<"\t"<<#y<<" :: "<<y<<"\n";
#define fr freopen("input.txt","r",stdin);
#define fw freopen("output.txt","w",stdout);
typedef long long int ll;
using namespace std;
int main()
{
    ll n,i;
    sl(n);
    if(n==0)
    {
        puts("0");
        return 0;
    }
    vector<int>ans;
    while(n!=0)
    {
        if(n%2==0)
        {
            ans.pb(0);
            n/=-2;
        }
        else if(n<0)
        {
            ans.pb(1);
            n/=-2;
            n++;
        }
        else
        {
            ans.pb(1);
            n/=-2;
        }
    }
    for(i=ans.size()-1;i>=0;i--)
        pf("%d",ans[i]);
}