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
# sys.stdin = open('input.txt', 'r')


class WayToZero:
    def __init__(self, n):
        self.n = n
        self.data = []
        self.output = []
        self.read()

    def read(self):
        for i in range(self.n):
            temp = [int(x) for x in input().split(' ')]
            self.data.append(temp)

    @staticmethod
    def calculate(x, k):
        result = -1
        while x != 0:
            modulo = x % k
            x //= k
            result += 1 + modulo

        return result

    def calculate_all(self):
        for i in range(self.n):
            self.output.append(self.calculate(self.data[i][0], self.data[i][1]))

    def __repr__(self):
        return '\n'.join([str(x) for x in self.output])

def main():
    way = WayToZero(int(input()))
    way.calculate_all()
    print(way)


main()
