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.
'''input
2
59 3
1000000000000000000 10000000000000
'''
# its just a simulation
from sys import stdin, setrecursionlimit
import heapq
import collections
import math

def brute(num, k, dp):
	steps = 0
	while True:
		if num == 0:
			break

		if num % k == 0:
			num = num//k
			steps += 1
			continue
		else:
			t = num % k
			num = num  - t
			steps += t

	return steps

# main starts
t = int(stdin.readline().strip())
for _ in range(t):
	dp = dict()
	n, k = list(map(int, stdin.readline().split()))
	print(brute(n, k, dp))
