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>
#define FAST ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define dofloat cout<<fixed<<setprecision(8)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define bitcount __builtin_popcount
#define all(vec) vec.begin(),vec.end()
#define rall(vec) vec.rbegin(),vec.rend()
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<long long> vl;
typedef pair<long long,long long> pll;
typedef vector<pair<long long,long long> > vll;
typedef vector<pair<int,int> > vii;
typedef vector<int> vi;
typedef pair<int,int> ii;
const long long MOD=1000000007;
const long long MAX=100005;
const long double PI=3.14159265359;
const long double G=9.807;
const long long INF=1e18;
const long double EPS=1e-6;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
bool isprime(ll a){
  if(a==2)return 1;
  if(!(a&1))return 0;
  for(ll i=3;i*i<=a;i+=2)
    if(a%i==0)return 0;
  return 1;
}
ll mpow(ll base,ll exponent,ll modulus){
  if(modulus==1)return 0;
  long long result = 1;
  base=base%modulus;
  while(exponent){
    if(exponent%2 == 1)
    result=(result*base)%modulus;
    exponent=exponent>>1;
    base=(base*base)%modulus;
  }
  return result;
}
ll minv(ll a,ll mod){
  ll _gcd=gcd(a,mod);
  assert(_gcd==1);
  return mpow(a,mod-2,mod);
}


/*
ll ncr(ll N,ll K){
  if(N<K)return 0;
  if(K==0)return 1;
  if(N==0)return 0;

  ll den=1;
  for(ll i=1;i<=K;i++)den*=i;
  ll num=1;
  while(K--){
    num*=N;
    if(num%den==0){
      num/=den;
      den=1;
    }
    N--;
  }

  return num;
}
*/
ll n,k;
int main(){FAST;
cin>>n>>k;
ll x = n;

if(n == 0){
  cout<<1;
  return 0;
}
ll ans = 0;
while(x){
  x/=k;
  ans++;
}
cout<<ans;
return 0;
}