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.
# AC
import sys


class Main:
    def __init__(self):
        self.buff = None
        self.index = 0

    def next(self):
        if self.buff is None or self.index == len(self.buff):
            self.buff = self.next_line()
            self.index = 0
        val = self.buff[self.index]
        self.index += 1
        return val

    def next_line(self):
        return sys.stdin.readline().split()

    def next_ints(self):
        return [int(x) for x in sys.stdin.readline().split()]

    def next_int(self):
        return int(self.next())

    def solve(self):
        ca = self.next_int()
        for _ in range(ca):
            n, k = self.next_ints()
            stp = 0
            while n > 0:
                if n % k == 0:
                    n //= k
                    stp += 1
                else:
                    stp += n % k
                    n -= n % k
            print(stp)


if __name__ == '__main__':
    Main().solve()
