You are given an integer n and an integer k.

In one step you can do one of the following moves: 

  * decrease n by 1; 
  * divide n by k if n is divisible by k. 



For example, if n = 27 and k = 3 you can do the following steps: 27 → 26 → 25 → 24 → 8 → 7 → 6 → 2 → 1 → 0.

You are asked to calculate the minimum number of steps to reach 0 from n. 

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of queries.

The only line of each query contains two integers n and k (1 ≤ n ≤ 10^{18}, 2 ≤ k ≤ 10^{18}).

Output

For each query print the minimum number of steps to reach 0 from n in single line. 

Example

Input


2
59 3
1000000000000000000 10


Output


8
19

Note

Steps for the first test case are: 59 → 58 → 57 → 19 → 18 → 6 → 2 → 1 → 0.

In the second test case you have to divide n by k 18 times and then decrease n by 1.
import java.util.Scanner;

public class Codeforce_A_FromHerotoZero {
    private static final Scanner SCANNER = new Scanner(System.in);

    public static void main(String args[]) {

        int queries = SCANNER.nextInt();

        for (int k = 0; k < queries; k++) {

            long integer_input = SCANNER.nextLong();
            long divisor = SCANNER.nextLong();

            long step = 0;
            long input = integer_input / divisor;

            if(integer_input < divisor){
                step = integer_input;
            }
            else{
                step = Calculate(integer_input, divisor, step);
            }

            System.out.printf("%d\n",step);
        }

        SCANNER.close();
    }

    public static long Calculate(long integer_input, long divisor, long step) {
        long remainder =(long) Math.floor(integer_input % divisor);

        integer_input -= remainder;
        step += remainder;

        if (integer_input == 0) {
            return step;
        }

        integer_input = (long) integer_input / divisor;
        step += 1;

        return Calculate(integer_input, divisor, step);
    }
}

// do a thing once for all not to separate ; recursive faster than for at the cost of memory