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

#### Input:

148 3

#### Output:

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 { //your code here 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]); } }

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