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>
template <class T>
std::vector<T> ReadVector(size_t size) {
  std::vector<T> vector(size);
  for (auto& element : vector) {
    std::cin >> element;
  }
  return vector;
}
int64_t divisor(int64_t x) {
  for (int64_t d = 2; d * d <= x; ++d) {
    while (x % d == 0) {
      x /= d;
      return d;
    }
  }
  return x;
}
void YES(bool yes) {
  if (yes) {
    std::cout << "YES\n";
  } else {
    std::cout << "NO\n";
  }
}
class Solver {
 public:
  std::vector<std::vector<int64_t>> v;
  Solver() {}
  void solve() {
    int64_t t;
    std::cin >> t;
    while (t--) {
      solve_one();
    }
  }
  void fail() { std::cout << -1 << std::endl; }
  void solve_one() {
    int64_t n, k;
    std::cin >> n >> k;
    int64_t count = 0;
    while (n > 0) {
      if (n % k != 0) {
        count += n % k;
        n -= n % k;
      } else {
        count += 1;
        n /= k;
      }
    }
    std::cout << count << std::endl;
  }
};
int main() {
  Solver solver;
  solver.solve();
  return 0;
}
