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.
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;
import java.util.Scanner;
import java.math.BigInteger; 

/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
   int t;
        Scanner sc = new Scanner(System.in);
        t = sc.nextInt();
        while(t>0)
        {
        BigInteger n,k,a,b,counta,c;
        a=new BigInteger("0");
        b=new BigInteger("1");
        counta=new BigInteger("0");
        n = sc.nextBigInteger();
        k = sc.nextBigInteger();
        while(n.compareTo(a) > 0)
        {
            c=n.remainder(k);
            if(c.compareTo(a) == 0)
                {
                    n=n.divide(k);
                    counta=counta.add(b);
                    
                }
            else
            {
                n=n.subtract(c);
                counta=counta.add(c);
            }
        }
        
        System.out.println(counta);
        t--;
        }
        sc.close();
	}
}
