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;
using llint = long long;
using pint = pair<int, int>;
const int INF = 1 << 28;
int t;
vector<llint> n;
vector<llint> k;
vector<llint> ans;
int main() {
  cin >> t;
  n.resize(t);
  k.resize(t);
  ans.resize(t);
  for (int i = 0; i < t; ++i) {
    cin >> n.at(i) >> k.at(i);
  }
  for (int i = 0; i < t; ++i) {
    llint cn = n.at(i);
    llint ck = k.at(i);
    while (cn != 0) {
      if (cn % ck == 0) {
        ++ans.at(i);
        cn /= ck;
      } else {
        ans.at(i) += cn - (cn / ck) * ck;
        cn = (cn / ck) * ck;
      }
    }
  }
  for (int i = 0; i < t; ++i) {
    cout << ans.at(i) << endl;
  }
  return 0;
}
