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 <bits/stdc++.h>
using namespace std;
using LL = long long int;
#define incID(i, l, r) for(int i = (l)    ; i <  (r); ++i)
#define decID(i, l, r) for(int i = (r) - 1; i >= (l); --i)
#define incII(i, l, r) for(int i = (l)    ; i <= (r); ++i)
#define decII(i, l, r) for(int i = (r)    ; i >= (l); --i)
#define inc(i, n)  incID(i, 0, n)
#define dec(i, n)  decID(i, 0, n)
#define inc1(i, n) incII(i, 1, n)
#define dec1(i, n) decII(i, 1, n)
#define inID(v, l, r) ((l) <= (v) && (v) <  (r))
#define inII(v, l, r) ((l) <= (v) && (v) <= (r))
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define FI first
#define SE second
#define FR front()
#define BA back()
#define ALL(v) v.begin(), v.end()
#define RALL(v) v.rbegin(), v.rend()
auto setmin   = [](auto & a, auto b) { return (b <  a ? a = b, true : false); };
auto setmax   = [](auto & a, auto b) { return (b >  a ? a = b, true : false); };
auto setmineq = [](auto & a, auto b) { return (b <= a ? a = b, true : false); };
auto setmaxeq = [](auto & a, auto b) { return (b >= a ? a = b, true : false); };
#define SI(v) static_cast<int>(v.size())
#define RF(e, v) for(auto & e: v)
#define until(e) while(! (e))
#define if_not(e) if(! (e))
#define ef else if
#define UR assert(false)
#define IN(T, ...) T __VA_ARGS__; IN_(__VA_ARGS__);
void IN_() { };
template<typename T, typename ... U> void IN_(T & a, U & ... b) { cin >> a; IN_(b ...); };
template<typename T> void OUT(T && a) { cout << a << endl; }
template<typename T, typename ... U> void OUT(T && a, U && ... b) { cout << a << " "; OUT(b ...); }

// ---- ----

int main() {
	IN(int, n, k);
	int ans = 0;
	while(n) { n /= k; ans++; }
	OUT(ans);
}