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.
'''
# Sample code to perform I/O:

name = input()                  # Reading input from STDIN
print('Hi, %s.' % name)         # Writing output to STDOUT

# Warning: Printing unwanted or ill-formatted data to output will cause the test cases to fail
'''

######################################################################
# Write your code here
import sys
#import resource
#resource.setrlimit(resource.RLIMIT_STACK, [0x10000000, resource.RLIM_INFINITY])
#https://stackoverflow.com/questions/3323001/what-is-the-maximum-recursion-depth-in-python-and-how-to-increase-it
sys.setrecursionlimit(0x100000)
# Write your code here
RI = lambda : [int(x) for x in input().strip().split()]
from collections import defaultdict as df
import heapq 
#heapq.heapify(li) heappush(li,4) heappop(li)
import random
#random.shuffle(list)
infinite = float('inf')
######################################################################3
t=int(input())

for i in range(t):
    n,k=RI()

    count=0
    while(n>0):
        temp=n%k
        count+=temp
        n-=temp
        n=n//k
        if(n!=0):
            count+=1

    print(int(count))
