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;
template <typename T>
inline void _read(T& x) {
  cin >> x;
}
void R() {}
template <typename T, typename... U>
void R(T& head, U&... tail) {
  _read(head);
  R(tail...);
}
template <typename T>
inline void _write(const T& x) {
  cout << x << ' ';
}
void W() { cout << endl; }
template <typename T, typename... U>
void W(const T& head, const U&... tail) {
  _write(head);
  W(tail...);
}
template <typename T>
void debug(vector<T> V) {
  for (auto x : V) {
    cout << x << " ";
  }
  cout << endl;
}
void go();
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  go();
  return 0;
}
void go() {
  int t;
  R(t);
  while (t--) {
    long long a, b;
    R(a, b);
    long long ans = 0;
    while (a) {
      if (a % b == 0) {
        ans++;
        a /= b;
      } else {
        long long w = a % b;
        ans += w;
        a -= w;
      }
    }
    W(ans);
  }
}
