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;
long long int ma = (long long int)(pow(2, 63));
map<long long int, long long int> mp;
void func1(long long int n, long long int k, long long int st) {
  long long int v = n % k;
  n = n - v;
  st = st + v;
  if (n == 0) {
    if (st < ma) ma = st;
    return;
  } else {
    long long int u = n / k;
    if (u <= k) {
      mp[u] = st + 1;
      func1(u, k, st + 1);
    } else {
      if (mp.find(u) != mp.end() && mp[u] > (st + 1)) {
        mp[u] = st + 1;
        func1(u, k, st + 1);
      } else if (mp.find(u) == mp.end())
        func1(u, k, st + 1);
      long long int z = u % k;
      u = u - z;
      if (mp.find(u) != mp.end() && mp[u] > (st + 1 + (n - (u * k)))) {
        mp[u] = st + 1 + (n - (u * k));
        func1(u, k, st + 1 + (n - (u * k)));
      }
    }
  }
}
int main() {
  long long int i, j, t, n, k;
  cin >> t;
  for (i = 0; i < t; i++) {
    cin >> n >> k;
    long long int v = n % k;
    func1(n - v, k, v);
    cout << ma << endl;
    ma = (long long int)(pow(10, 18));
    mp.clear();
  }
  return 0;
}
