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 sys
def fastio():
	from io import StringIO
	from atexit import register
	global input
	sys.stdin = StringIO(sys.stdin.read())
	input = lambda : sys.stdin.readline().rstrip('\r\n')
	sys.stdout = StringIO()
	register(lambda : sys.__stdout__.write(sys.stdout.getvalue()))
fastio()

def debug(*var, sep = ' ', end = '\n'):
	print(*var, file=sys.stderr, end = end, sep = sep)

INF = 10**20
MOD = 10**9 + 7
I = lambda:list(map(int,input().split()))
from math import gcd
from math import ceil
from collections import defaultdict as dd, Counter
from bisect import bisect_left as bl, bisect_right as br

t, = I()
while t:
	t -= 1
	n, k = I()
	ans = 0
	while n != 0:
		if n % k == 0:
			n //= k
			ans += 1
		else:
			m = n % k
			ans += m
			n -= m
	print(ans)