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.*;
import java.lang.reflect.Array;
import java.util.*;

import static java.lang.Math.*;

public class Main {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");
    PrintWriter pw = new PrintWriter(System.out);

    int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    String next() throws IOException {
        if (!st.hasMoreTokens()) {
            st = new StringTokenizer(br.readLine());
        }
        return st.nextToken();
    }

    long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    public static void main(String[] args) throws IOException {
        new Main().run();
    }


    void run() throws IOException {
        int t;
        t = nextInt();
        for (int i = 0; i < t; i++) {
            long a = nextLong();
            long b = nextLong();
            long ans = 0;
            while (a != 0) {
                if (a % b != 0) {
                    ans += a % b;
                    a -= a % b;
                } else {
                    a /= b;
                    ans++;
                }
            }
            pw.println(ans);
        }

        pw.close();
    }
}