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;
int nxt() {
  int x;
  cin >> x;
  return x;
}
long long int nxtl() {
  long long int x;
  cin >> x;
  return x;
}
void SieveOfEratosthenes(int n, vector<int> &primes) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (int p = 2; p <= n; p++)
    if (prime[p]) primes.push_back(p);
}
long long int max(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int power(long long int x, long long int y, long long int mod) {
  long long int temp;
  if (y == 0) return 1;
  temp = power(x, y / 2, mod);
  if (y % 2 == 0)
    return (temp * temp) % mod;
  else
    return (((x * temp) % mod) * temp) % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = nxt();
  while (t--) {
    long long int n = nxtl(), k = nxtl();
    long long int x = 0;
    while (n > 0) {
      long long int r = n % k;
      n -= r;
      x += r;
      if (n > 0) {
        n /= k;
        x++;
      }
    }
    cout << x << endl;
  }
  return 0;
}
