## Recursive Digit Sum Problem

We define super digit of an integer n using the following rules:
Given an integer, we need to find the super digit of the integer n.
If n has only 1 digit, then its super digit is n.
Otherwise, the super digit of n is equal to the super digit of the sum of the digits of n.

superDigit has the following parameter(s):

string n: a string representation of an integer
int k: the times to concatenate n to make p.

#### Input

The first line contains two space separated integers, n and k.

#### Constraints:

1 <= n <= 10^100000
1 <= k <= 100000

#### Output

In a new line, print the the super digit of n repeated k times.

### Example

```148 3
```

```3
```

#### Explanation:

```Here n=148 and k=3 , so p=148148148.

super_digit(p) = super_digit(148148148)
= super_digit(1+4+8+1+4+8+1+4+8)
= super_digit(39)
= super_digit(3+9)
= super_digit(12)
= super_digit(1+2)
= super_digit(3)
= 3```

## Solution of Recursive Digit Sum Problem :–

```import java.util.*;
import java.lang.*;
import java.io.*;

public class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
for(int j=1;j<=t;j++)
{
int n=sc.nextInt();
int[] arr=new int[n];
for(int i=0;i<n;i++)
{
arr[i]=sc.nextInt();
}
System.out.print("Case "+j+": "+sum(arr,n)+"\n");
}
}
public static int sum(int[] arr,int n)
{
if(n<=0)
{
return 0;
}
return (sum(arr,n-1)+arr[n-1]);
}
}```