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.
#include <bits/stdc++.h>
using namespace std;
const int LIMIT = 1e5;
int ans[LIMIT];
int n;
int main() {
  cin >> n;
  long long m[n], k[n];
  for (int i = 1; i <= n; i++) {
    cin >> m[i] >> k[i];
  }
  for (int i = 1; i <= n; i++) {
    long long steps = 0;
    while (m[i] >= k[i]) {
      steps += (m[i] % k[i]);
      m[i] /= k[i];
      steps++;
    }
    steps += m[i];
    cout << steps << endl;
  }
  return 0;
}
