//10+9+9*10+9*10+9*10*10...
//9*10^(ceil(n/2)-1)
//9^ceil(n/2)
#include <iostream>
#include <cmath>                                                     
#include <cstring>

using namespace std;

typedef long long LL;

LL x;

int main()
{
    cin>>x;
    if(x<=10) cout<<x-1;
    else
    {  
        x-=10;
        LL n=2;
        LL temp=9;
        // while(temp<=x) //优质写法
        // {
        //     x-=temp;
        //     n++;
        //     if(n%2==1)
        //         temp*=10;
        // }
        while(9ll*pow(10,ceil((double)n/2)-1)<=x)
        {
            //LL temp=9ll*pow(10,ceil((double)n/2)-1);//pow(产出为浮点数，要强制转化为long long,否则会溢出)
            x-=9ll*(LL)pow(10,ceil((double)n/2)-1);
            n++;
        }
        //cout<<x<<endl;
        //cout<<x-9<<endl;
        LL ans=pow(10,ceil((double)n/2)-1);
        ans+=x-1;
        //cout<<ans<<endl;
        string res=to_string(ans);
        if(n%2==0)
        {
            cout<<res;
            for(int i=res.size()-1;i>=0;i--) cout<<res[i];
        }
        else
        {
            cout<<res;
            for(int i=res.size()-2;i>=0;i--) cout<<res[i];
        }
    }
}