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)

#include <iostream>
#include <vector>
#include <limits.h>
#include <algorithm>
#include <string>
#include <math.h>
#include <limits.h>
#include <queue>
#include <map>
#include <set>
#include <iomanip>
#include <bitset>
#include <cassert>
#include <random>
#include <functional>
#include <stack>
#include <iomanip>
#include <cassert>
//#include <boost/multiprecision/cpp_int.hpp>
#include <complex>
#include <cstdio>
#include <list>

//< in.txt > out.txt
using namespace std;
//std::ios::sync_with_stdio(false);
//std::cin.tie(0);
const long long MOD = 1e9 + 7;
typedef long long LL;
typedef long double LD;
typedef pair<LL, LL> PLL;
typedef pair<LD, LL> PDL;
typedef pair<LD, LD> PDD;
typedef vector<LL> VLL;
typedef vector<VLL> VVLL;
//typedef boost::multiprecision::cpp_int bigint;

template<class T>
void in(T& x) {
	cin >> x;
}

template<class T1,class T2>
void in(pair<T1, T2>& p) {
	in(p.first);
	in(p.second);
}

template<class T>
void in(vector<T>& v,LL st=-1,LL en=-1) {
	if (st == -1) {
		st = 0;
		en = v.size() - 1;
	}
	for (LL n = st; n <= en; n++) {
		in(v[n]);
	}
}

int main() {
	std::ios::sync_with_stdio(false);
	std::cin.tie(0);
	LL N, K;
	cin >> N >> K;
	LL temp = K;
	for (LL n = 1;; n++) {
		if (temp > N) {
			cout << n << "\n";
			return 0;
		}
		temp *= K;
	}
	return 0;
}