cf-431D

/* the number of ones between n+1 to 2n increases with n...so given a certain number m  we can find which n and 2n have m numbers with exactly k digits one by binary search and digit dp */


/*********************** let the play begin() ***********************************/

vl v1,v2;ll dp[70][2][2][70],m,k;int lim;

ll giveres(int pos,int low,int high,ll pk)
{
    if(pk>k)return 0ll;
    if(pos==lim)
    {
        if(pk==k)return 1ll;
        return 0ll;
    }

    ll &ret = dp[pos][low][high][pk];

    if(ret!=-1)return ret;

    ret = 0;
    int h = 1,l = 0;
    if(high)h = v2[pos];
    if(low)l = v1[pos];

    for(int i = l;i<=h;i++)
    {
        ret+=giveres(pos+1,low&(i==l),high&(i==h),pk+i);
    }

    return ret;
}

int main()
{
    booster()
    ///read("input.txt");


    cin>>m>>k;

    ll low = 1,high = 1e18,mid;
    ll ans = -1;
    while(low<=high)
    {
        mid = (low+high)>>1;
        ll x = mid;v1.clear();v2.clear();
        ll y = 2*x;
        x++;
        while(x)
        {
            v1.pb(x%2);
            x/=2;
        }
        while(y)
        {
            v2.pb(y%2);
            y/=2;
        }
        while(v1.size()<v2.size())v1.pb(0);
        reverse(all(v1));
        reverse(all(v2));
        lim = v1.size();
        mem(dp,-1);
        if(giveres(0,1,1,0)>=m)
        {
            ans = mid;
            high = mid - 1;
        }
        else low = mid + 1;
        ///cout<<mid<<" "<<2*mid<<" "<<giveres(0,1,1,0)<<endl;

    }

    cout<<ans;

    return 0;
}