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 K {
	Scanner t = new Scanner(System.in);
	long k, n;
	String acumulado;
	
	public void captura() {
		String s;
		String[] nom;		
		int vueltas;
				
		vueltas = Integer.parseInt(t.nextLine());
		if(vueltas>=1 && vueltas<=100) {				
			while(vueltas>0) {			
				s =t.nextLine();
				nom = s.split(" ");
				n = Long.parseLong(nom[0]);
				k = Long.parseLong(nom[1]);			
				if(acumulado == null)
					acumulado = ope();
				else
					acumulado += ope();
				vueltas= vueltas-1;
			}	
			System.out.println(acumulado);			
		}
	}
	

	public String ope() {
		long contador=0;
		while(n>0) {
			if(k==1) {				
				contador=n;
			}else if(n%k==0) {
				n = n/k;
				contador++;
			}
			else {
				contador+= (n%k);
				n=(n-(n%k));				
			} 
		}
		return (contador+"\n");
	}	
	
	
	public static void main(String[] args) {
		K kp = new K();
		kp.captura();		
	}

}
//YA SABEMOS QUE NO CUENTA XD JEJEJEJE
				  	 	 	    		 	   	  			 		