Problem:
Score : 200 points

Problem Statement

Given is an integer N. Find the number of digits that N has in base K.

Notes

For information on base-K representation, see Positional notation - Wikipedia.

Constraints

All values in input are integers.

1 \leq N \leq 10^9

2 \leq K \leq 10

Input

Input is given from Standard Input in the following format:

N K

Output

Print the number of digits that N has in base K.

Sample Input 1

11 2

Sample Output 1

4

In binary, 11 is represented as 1011.

Sample Input 2

1010101 10

Sample Output 2

7

Sample Input 3

314159265 3

Sample Output 3

18

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <algorithm>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <functional>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <vector>
using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const int dx[4] = {-1, 0, 0, 1};
const int dy[4] = {0, -1, 1, 0};

// Self settings
// clang-format off
#define MAX_N 100000
#define REP(i, N) for (int i = 0; i < (int)(N); ++i)
#define SLN(i,N) (i == N-1 ? "\n" : " ")
ll fact(ll n) { ll res = 1; for(ll i=2;i<=n;++i) res = res * i; return res;}
ll nCr(ll n, ll r) {return (fact(n)/fact(n-r)*fact(r)) ;}
ll gcd(ll a,ll b){if(b==0)return a;return gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
const ll MOD = 1e9+7;
const ll INF = 1LL << 60;
const int inf = 1000100011;
// clang-format on

int main(void) {
  cin.tie(0);
  ios::sync_with_stdio(false);
  ll N, K;
  cin >> N >> K;

  int cnt = 0;
  while (N > K) {
    N = ceil((double)N / K);
    cnt++;
  }
  cnt++;
  cout << cnt << endl;

  return 0;
}