#include<bits/stdc++.h>
using namespace std;
#define endl "\n"
#define debug(x) cout<<#x<<": "<<x<<endl
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
using ll=long long;
using ull=unsigned long long;
double a,b;
double eps =1e-4;
void solve(){
    cin>>a>>b;
    while(abs(a-b)>=eps){
        if(a-b>=0){
            a=a*0.5;
            cout<<1;
        }else{
            cout<<2;
            a=(a+1)*0.5;
        }
    }
    cout<<endl;

}
signed main(){
    ios::sync_with_stdio(0);
    cout.tie(0);
    cin.tie(0);
    int t=1;
    while(t--){
        solve();
    }
}
/*
贡献法
正难则反
数小状压
关系连边
拆位
广义单调性
最长转最短
*/