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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Main {
	public static long index = 0;
	public static void main(String[] args) throws Exception {
		BufferedReader in = new BufferedReader (new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
		int  casos = Integer.parseInt(in.readLine());
		String linea;
		for (int i = 0; i < casos; i++) {
			linea = in.readLine();
			StringTokenizer st = new StringTokenizer(linea);
			long n = Long.parseLong(st.nextToken());
			long k = Long.parseLong(st.nextToken());
			if(k == 1) index = n;
			else kMethod(n, k);
			out.println(index);
			index = 0;
		}
		out.close();
	}
	
	public static void  kMethod (long a, long b) {
		while(a > 0) {
			if(a%b == 0) {
				a = a/b;
				index++;
			}
			else {
				long dif = a % b;
				a -= dif;
				index += dif;
			}
		}
	}

}
 		  				 	 	  		 		   		 					