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;
#define ll long long int
#define ld long double
#define pb emplace_back
#define mp make_pair
#define f(i,a,n) for(int i=a ; i<n ; i++)
#define F first
#define S second
#define vll vector<ll>
#define fast ios:: sync_with_stdio(false),cin.tie(0);
#define COUT cout<<fixed<<setprecision(12)
#define SET set<ll,greater<ll>>gquiz
#define ITERATOR set<ll,greater<ll>> :: iterator itr
#define V vector<ll>vector
#define sz size()
#define A vector<pair<ll,ll>>vect
ll N=100000,M=100000;
#define B vect.push_back(make_pair(N,M));
ll MOD=1000000007;
 
ll gcd(ll a,ll b) 
{ 
    if (a == 0) 
        return b; 
    return gcd(b % a, a); 
} 
ll ncr(ll n,ll r)
{
	if(n==1 || r==0)
	return 1;
	if(r>n/2)
	r=n-r;
	ll ans=1;
	for(ll i=1;i<=r;i++)
	{
		ans=((ans * n )/i);
		n--;
	}
	return ans;
}
int main()
{
	ll n,k;
	cin>>n>>k;
	ll cnt=0;
	while(n>0)
	{
		n=n/k;
		cnt++;
	}
	cout<<cnt;
}