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: Accepted)

/*
このコード、と～おれ!
Be accepted!
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					*/

#include <iostream>
#include <vector>
#include <array>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <climits>
#include <tuple>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <cmath>
#include <algorithm>
#include <memory>
#include <iomanip>
#include <type_traits>

#define rep(i, n) for(Int (i) = 0; (i) < (n); ++(i))
#define all(n) std::begin(n), std::end(n)
#define debug(var) std::cout << #var << ": " << var << "\n"
using Int = int64_t;
constexpr Int INF = INT_MAX;
constexpr Int mod = 1e9 + 7;

double log(double base, double value) {
	return std::log2(value) / std::log2(base);
}


void init() {
	std::cin.tie(nullptr);
	std::ios_base::sync_with_stdio(false);
	std::cout << std::fixed << std::setprecision(15);
}

void calc() {

	Int N, K; std::cin >> N >> K;
	Int ans = 0;
	while (N > 0) {
		N /= K;
		ans++;
	}

	std::cout << ans;

}


int main() {

	init();
	calc();

	return 0;
}