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;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;typedef vector<ll> vll;
#define fill(a) memset(a, 0, sizeof (a))
#define all(v) (v).begin(),(v).end()
#define sp(k) cout<<setprecision(k)<<fixed;
#define rep(i,a) for(int i=0; i<a;i++)
#define rep1(i,a,b) for(int i=(a);i<=(b);++i)
#define irep(i,b,a) for(int i=(b);i>=(a);--i)
#define minv(v) *min_element(v.begin(),v.end())
#define maxv(v) *max_element(v.begin(),v.end())
#define each(it,s) for(auto it = s.begin(); it != s.end(); ++it)
#define unq(v) sort(all(v)),(v).erase(unique((v).begin(),(v).end()),(v).end())
#define getpos(v,x) (lower_bound(v.begin(),v.end(),x)-v.begin())
#define _ios ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define ff first
#define ss second
#define pb push_back
#define sz(x) (int)x.size()

const double pi = acos(-1.0);
const int mod = 1000000007; // (int)1e9+7
const int maxn = 2000100;
void solve(ll n, ll base)
{
    int dig = (int)(floor( log(n) /  log(base)) + 1);
    cout<<dig<<endl;
}
int main(){
        _ios;
        ll n,k;
        cin>>n>>k;
        solve(n,k);




return 0;
}