#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define op(x) ((x&1)?x+1:x-1)
#define odd(x) (x&1)
#define even(x) (!odd(x))
#define lc(x) (x<<1)
#define rc(x) (lc(x)|1)
#define lowbit(x) (x&-x)
#define mp(x,y) make_pair(x,y)
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
using namespace std;
const int MAXN = (1<<20)+10;
typedef array<int,2> pr;

int n,k,U;

vector<int> e[MAXN];
int st[MAXN],cur[MAXN],top;

int ans[MAXN],tot;

void dfs(int u){
    for(int& j=cur[u];j<e[u].size();){
        int v = e[u][j]; j++;
        dfs(v);
    }
    st[++top] = u;
}

namespace T{
    int cyc[MAXN],len,vis[MAXN];

    int to[MAXN];

    void solve(){
        //先变成2^k意义下的欧拉路
        U<<=1;
        for(int i=1;i<top;i++){
            cyc[++len] = (st[i]<<1) | (st[i+1]&1);
            cyc[len] %= U;
        }
        U<<=1;
        //for(int i=1;i<=len;i++)cout<<cyc[i]<<" "; cout<<endl;
        //再变成2^(k+1)意义下的
        for(int i=len-1;i;i--){
            cyc[i+1] = (cyc[i]<<1) | (cyc[i+1]&1);
            cyc[i+1]%=U;
        }
        cyc[1] = (cyc[len]<<1) | (cyc[1]&1); cyc[1]%=U;
        //for(int i=1;i<=len;i++)cout<<cyc[i]<<" "; exit(0);

        for(int i=1;i<=len;i++)vis[cyc[i]] = 1;
        for(int i=0;i<U;i++)to[i] = -1;
        for(int i=1;i<len;i++)to[cyc[i]] = cyc[i+1]; to[cyc[len]] = cyc[1];
        for(int i=0;i<U;i++)if(to[i]==-1){
            int j = (i+(U/2))%U; 
            //printf("i:%d j:%d\n",i,j);
            assert(to[j] != -1);
            //printf("i:%d j:%d(%d)\n",i,j,to[j]);
            to[i] = (i<<1) | !(to[j]&1); to[i] %= U;
            //printf("get %d\n",to[i]);
        }
        //for(int i=0;i<U;i++)cout<<to[i]<<" "; exit(0);
        //for(int j=1;j<=len;j++)cout<<cyc[j]<<" "; cout<<endl;
        for(int i=0;i<U;i++)if(!vis[i]){ //找到一个新环
            int u = i,cnt = 0;
            while(!vis[u])vis[u] = 1,u = to[u],cnt++;
            int uu = (u+(U/2))%U; swap(to[u],to[uu]);
            //printf("with u:%d uu:%d\n",u,uu);
            //printf("%d : %d %d\n",i,cnt,to[i]);
            len += cnt;
            if(len < n-k)continue;
            len = tot = 0;
            for(int j=1;j<=n-k;j++)cyc[++len] = u,u = to[u];

            //cout<<"done path : ";
            //for(int j=1;j<=len;j++)cout<<cyc[j]<<" "; cout<<endl;
            
            for(int j=k;j>=0;j--)ans[++tot] = cyc[1]>>j&1;
            for(int j=1;j<len;j++)ans[++tot] = cyc[j+1]&1;
            for(int i=1;i<=n;i++)cout<<ans[i];
            exit(0);
        }
        assert(0);
    }
}

int main(){
    //freopen("string.in","r",stdin);
    //freopen("string.out","w",stdout);

    cin>>n;
    if(n==1)return (cout<<"0\n"),0;
    if(n==2)return (cout<<"01\n"),0;
    if(n==3)return (cout<<"010\n"),0;
    if(n==4)return (cout<<"0110\n"),0;
    k = 1; while((1<<k)+k-1 <= n)k++; k--;

    U = 1<<(k-1);
    for(int i=0;i<U;i++){
        e[i].push_back((i<<1)%U);
        e[i].push_back(((i<<1)|1)%U);
    }

    dfs(0); reverse(st+1,st+1+top);

    tot = k-1;
    for(int i=1;i<top;i++){
        if(((st[i]<<1)%U) == st[i+1])ans[++tot] = 0;
        else ans[++tot] = 1;
    }
    if((1<<k)+k-1 < n)T::solve();

    for(int i=1;i<=tot;i++)cout<<ans[i];

    return 0;
}