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: Accepted)

#pragma region Header

#include <iostream>
#include <iomanip>
#include <vector>
#include <unordered_map>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <array>
#include <tuple>
#include <string>
#include <algorithm>
#include <functional>
#include <cmath>
#include <numeric>
#include <bitset>
#include <cstdint>
#include <stdexcept>
using namespace std;

// 整数型

using i32 = int32_t;
using i64 = int64_t;
using u32 = uint32_t;
using u64 = uint64_t;

// [0,n)を昇順/降順でループ

#define rep(i, n) for (int i = 0; i < static_cast<int>(n); ++i)
#define repd(i, n) for (int i = static_cast<int>(n) - 1; i >= 0; --i)

// [a,b)を昇順/降順でループ

#define FOR(i, a, b) for (int i = static_cast<int>(a); i < static_cast<int>(b); ++i)
#define FORD(i, a, b) for (int i = static_cast<int>(b) - 1; i >= static_cast<int>(a); --i)

template <typename T>
int SZ(T x)
{
    return static_cast<int>(x.size());
}

#pragma endregion

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(15);

    int n, k;
    cin >> n >> k;
    int ans = 0;
    while (n > 0)
    {
        n /= k;
        ++ans;
    }
    cout << ans << endl;

    return 0;
}