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 code.forces;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.stream.Stream;

public class Solution {
	private static final BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
	
	public static void main(String[] args) throws IOException {
		int t = Integer.parseInt(reader.readLine());
		
		for(int i = 0; i < t; i++) {
			solve();
		}
		
	}

	private static void solve() throws IOException {
		long nk[] = readLongs(reader);
		
		long n = nk[0], k = nk[1];
		
		long count = 0;
		
		while(n > 0) {
			if(n % k == 0) {
				count++;
				n = n / k;
			} else {
				count += n % k;
				n -= (n % k);
			}
		}
		System.out.println(count);
	}
	
	public static long[] readLongs(BufferedReader reader) throws IOException {
		return toLongArray(reader.readLine());
	}
	
	public static long[] toLongArray(String s) {
		return Stream.of(s.split("\\s")).mapToLong(Long::parseLong).toArray();
    }
	
}
