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 java.util.Scanner;

public class FromHeroToZero {

  public static void main(String[] args) {
    Scanner myScanner = new Scanner(System.in);
    int iNumQueries = myScanner.nextInt();
    long[] aMinSteps = new long[iNumQueries];

    for(int i = 0; i < iNumQueries; i++) {
      long n = myScanner.nextLong();
      long k = myScanner.nextLong();
      long lNumberOfSteps = 0;
      while(n != 0) {
        if(n % k == 0) {
          n /= k;
          lNumberOfSteps++;
        }
        else {
          //-- In order to prevent having to subtract 1's O(n) times,
          //-- we calculate the number of one to be subtracted in order
          //-- to reach the next divisor of k
          long lDistanceToNextDivisorOfK = n % k;
          n -= lDistanceToNextDivisorOfK;
          lNumberOfSteps += lDistanceToNextDivisorOfK;
        }
      }
      aMinSteps[i] = lNumberOfSteps;
    }
    for(int i = 0; i < aMinSteps.length; i++) {
      System.out.println(aMinSteps[i]);
    }
  }
}
