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 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i, n) repi(i, 0, n)
#define repi(i, a, b) for (int i = (int)(a) ; i < (int)(b) ; i++)


#define SZ(x) ((int)(x).size())


typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
struct Vec{
    double x, y;
};

template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
T min(T a, T b)
{
    if (a < b) return a;
    return b;
}

template<class T>
T max(T a, T b)
{
    if (a > b) return a;
    return b;
}

using namespace std;

int main()
{
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    // cout << fixed << setprecision(15);
    
    ll N, d = 0;
    int K;
    cin >> N >> K;

    while (N > pow(K, d)){
        d++;
    }
    cout << d-- << endl;

    return 0;
}