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 (GCC 5.4.1) (judge verdict: Wrong Answer)

/*
  kadai3_3 Digits
  unomi
*/
#include <stdio.h>
#include <math.h>

double log_n(double N, double K){
  // 底の変換公式を用いて、log_K(N)を計算
  return log(N)/log(K);
}

int main(void){
  double N=0, K=0;

  // N,K の入力を取得
  scanf("%lf %lf",&N, &K);

  // N = K^log_K(N) < K^(log_K(N)を切り上げた整数) 
  // → K進数において、Nはlog_K(N)を切り上げた整数桁
  printf("%f\n",(int)ceil(log_n(N,K)));

  return 0;
}