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)

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 

using namespace __gnu_pbds; 
using namespace std;
using   ll=long long;
using   ld=long double;

#define fast        	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define ordered_set 	tree<pll, null_type,less<pll>, rb_tree_tag,tree_order_statistics_node_update>

#define endl        	"\n"
#define pb          	push_back
#define mp          	make_pair
#define F           	first
#define S           	second
#define int             long long int

#define ALL(v)      	v.begin(),v.end()
#define ALLR(v)     	v.rbegin(),v.rend()
#define pii         	3.14159265358979323
#define inf     		LLONG_MAX
#define ones(x)     	__builtin_popcount(x)
#define fill(a,b)   	memset(a,b,sizeof(a))
#define mod 			1000000007

ll mod_pow(ll a,ll b) 
{
    ll res = 1;
    while(b)	
    {
        if(b&1) 	
        {
        	res=(res*a)%mod;
        }
        a=(a*a)%mod;
        b>>=1;
    }
    return res;
}

ll mod_inverse(ll a) 
{
	return mod_pow(a , mod-2);
}

void solve()
{
	int n , k;
	
	cin >> n >> k;
	int tot = 0;
	
	while(n > 0) {
		tot++;
		n /= k;
	}
	cout << tot;
}

signed main() {
    fast;
    
    int t = 1;
    
    //cin >> t;
    
    while(t--) {
        solve();
    }
 
    return 0;
}