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.*;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		HashSet<Long> hs = new HashSet<>();
		int T = sc.nextInt();
		
		for(int test_case = 1; test_case <= T; test_case++)
		{
			long N = sc.nextLong();
			long K = sc.nextLong();
			if(N < K)
			{
				System.out.println(N);
				continue;
			}
			if(K == 1)
			{
				System.out.println(N);
				continue;
			}
			
			PriorityQueue<Pair> pq = new PriorityQueue<>();
			pq.add(new Pair(N, 0));
			hs.clear();
			while(!pq.isEmpty())
			{
				Pair p = pq.poll();
				if(p.num == 0) 
				{
					System.out.println(p.cnt);
					break;
				}
				if(p.num % K == 0 && !hs.contains(p.num / K))
				{
					hs.add(p.num / K);
					pq.add(new Pair(p.num / K, p.cnt + 1));
				}
				else
				{
					long mod = p.num % K;
					hs.add(p.num - mod);
					pq.add(new Pair(p.num - mod, p.cnt + mod));
				}
			}
		}
	}
}
class Pair implements Comparable<Pair>{
	long num, cnt;
	Pair(long num, long cnt)
	{
		this.num = num;
		this.cnt = cnt;
	}
	@Override
	public int compareTo(Pair o) {
		return o.cnt > this.cnt ? -1 : 1;
	}
}