#define ll long long

int superDigit(string n, int k) {
    string tmp = n;
    
    int len = tmp.size();
    int res = 0;
    
    while (len > 1) {
        ll sum = 0ll; 
        
        for (int i = 0; i < len; i++) {
            sum += int(tmp[i]-'0');
        }
        sum *= k;
        k=1;
        
        res = sum;
        
        string next = "";
        while (sum) {
            next += char((sum%10) + '0');
            sum /= 10;
        }
        tmp = string(next.rbegin(), next.rend());
        
        len = next.size();
    }
    return res;
}

def superDigit(n, k):
   # Base case: If n is already a single-digit number, return it as the super digit
   if len(n) == 1:
      return int(n)
   # Calculate the sum of the digits of n multiplied by k
   added = sum(int(i) for i in n) * k
   # Recursive call: with the sum converted to string and k set to 1
   # since repeated multiplication is only needed once  
   return superDigit(str(added), 1)

public static int superDigit(string n, int k)
    {
        long digitSum(string str)
        {
            long sum = 0;
            foreach (char c in str)
            {
                sum += c - '0';
            }
            return sum;
        }

        long initialSum = digitSum(n) * k;

        while (initialSum >= 10)
        {
            initialSum = digitSum(initialSum.ToString());
        }

        return Convert.ToInt32(initialSum);

    }

def superDigit(n, k):
    return (sum(map(int, list(n))) * k) % 9 or 9

const sumDigits = (n) => String(n).split('').reduce((acc, cur) => acc + Number(cur), 0)
function superDigit(n, k) {
    if (n < 10) {
        const value = sumDigits(n * k)
        const output = value < 10 ? value : superDigit(value, 1)
        return output
    } else {
        const step = sumDigits(n)
        return superDigit(step, k)
    }
}