#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long n, k;
long long check(long long a[], long long b[], long long mid) {
  long long sum = 0;
  for (long long i = 0; i < n; i++) {
    if (mid * a[i] - b[i] > 0) {
      sum += (mid * a[i] - b[i]);
      if (sum > k) return 0;
    }
  }
  if (sum <= k)
    return 1;
  else
    return 0;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    long long cnt = 0;
    cin >> n >> k;
    if (k == 1) {
      cout << n << endl;
      continue;
    }
    while (n > 0) {
      if (n % k == 0) {
        n = n / k;
        cnt++;
      } else {
        long long x = n % k;
        cnt += x;
        n -= x;
      }
    }
    cout << cnt << endl;
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
using namespace std;
long long mod = 1e9 + 7;
long long n, k;
long long check(long long a[], long long b[], long long mid) {
  long long sum = 0;
  for (long long i = 0; i < n; i++) {
    if (mid * a[i] - b[i] > 0) {
      sum += (mid * a[i] - b[i]);
      if (sum > k) return 0;
    }
  }
  if (sum <= k)
    return 1;
  else
    return 0;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    long long cnt = 0;
    cin >> n >> k;
    if (k == 1) {
      cout << n << endl;
      continue;
    }
    while (n > 0) {
      if (n % k == 0) {
        n = n / k;
        cnt++;
      } else {
        long long x = n % k;
        cnt += x;
        n -= x;
      }
    }
    cout << cnt << endl;
  }
}

```