#include <bits/stdc++.h>
#pragma GCC optimize("-O2")
using namespace std;
void err(istream_iterator<string> it) { cerr << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << "\t";
  err(++it, args...);
}
template <typename T1, typename T2>
ostream& operator<<(ostream& c, pair<T1, T2>& v) {
  c << "(" << v.first << "," << v.second << ")";
  return c;
}
template <template <class...> class TT, class... T>
ostream& operator<<(ostream& out, TT<T...>& c) {
  out << "{ ";
  for (auto& x : c) out << x << " ";
  out << "}";
  return out;
}
const int LIM = 1e5 + 5, MOD = 1e9 + 7;
const long double EPS = 1e-9;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tests;
  cin >> tests;
  while (tests--) {
    long long n, k;
    cin >> n >> k;
    long long ans = 0;
    while (n > 0) {
      if (n % k != 0) {
        ans += n % k;
        n -= n % k;
      }
      n = n / k;
      ans++;
    }
    cout << ans - 1 << '\n';
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("-O2")
using namespace std;
void err(istream_iterator<string> it) { cerr << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << "\t";
  err(++it, args...);
}
template <typename T1, typename T2>
ostream& operator<<(ostream& c, pair<T1, T2>& v) {
  c << "(" << v.first << "," << v.second << ")";
  return c;
}
template <template <class...> class TT, class... T>
ostream& operator<<(ostream& out, TT<T...>& c) {
  out << "{ ";
  for (auto& x : c) out << x << " ";
  out << "}";
  return out;
}
const int LIM = 1e5 + 5, MOD = 1e9 + 7;
const long double EPS = 1e-9;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tests;
  cin >> tests;
  while (tests--) {
    long long n, k;
    cin >> n >> k;
    long long ans = 0;
    while (n > 0) {
      if (n % k != 0) {
        ans += n % k;
        n -= n % k;
      }
      n = n / k;
      ans++;
    }
    cout << ans - 1 << '\n';
  }
  return 0;
}

```