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)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;

#define rep(i,n) for(long long i=0;i<(long long)(n);i++)
#define REP(i,k,n) for(long long i=k;i<(long long)(n);i++)
#define all(a) a.begin(),a.end()
#define NP next_permutation
#define pb push_back
#define eb emplace_back
#define lb(v,k) (lower_bound(all(v),k)-v.begin())
#define ub(v,k) (upper_bound(all(v),k)-v.begin())
#define fi first
#define se second
typedef long long ll;
typedef pair<ll,ll> P;
typedef tuple<ll,ll,ll> PP;
typedef multiset<ll> S;
typedef priority_queue<ll> PQ;
typedef priority_queue<P,vector<P>,greater<P>> SPQ;
using vi=vector<ll>;
using vvi=vector<vi>;
using vvvi=vector<vvi>;

const ll inf=1001001001001001;
const int INF=1001001001;
const int mod=1000000007;
const double pi=3.14159265358979323846;

template<class T> bool chmin(T&a,T b){if(a>b){a=b; return true;}return false;}
template<class T> bool chmax(T&a,T b){if(a<b){a=b; return true;}return false;}
template<class T> void decimal(T a){cout<<fixed<<setprecision(a);}
template<class T> void out(T a){cout<<a<<endl;}
template<class T> void outp(T a){cout<<'('<<a.fi<<','<<a.se<<')'<<endl;}
template<class T> void outvp(T v){rep(i,v.size())outp(v[i]);}
template<class T> void outv(T v){rep(i,v.size()){if(i)cout<<' ';cout<<v[i];}cout<<endl;}
template<class T> void outvv(T v){rep(i,v.size()){rep(j,v[i].size()){if(j)cout<<' ';cout<<v[i][j];}cout<<endl;}}

void YesNo(bool a){if(a) out("Yes"); else out("No");}
void yesno(bool a){if(a) out("yes"); else out("no");}
void YESNO(bool a){if(a) out("YES"); else out("NO");}
ll GCD(ll a,ll b){ll c; while(b!=0){c=a%b; a=b; b=c;} return a;}
ll LCM(ll a,ll b){return(a/GCD(a,b))*(b/GCD(a,b))*GCD(a,b);}
ll POW(ll a,ll b){ll c=1; while(b>0){if(b&1){c=a * c%mod;} a=a * a%mod; b>>=1;} return c;}
vi calc(ll x){vi res; while(x>0){res.pb(x%10); x/=10;} reverse(all(res)); return res;}

void solve_A() {
  int n,r;
  cin>>n>>r;
  if(n<10) out(r+100*(10-n));
  else out(r);
}
void solve_B() {
  int n,k,ans=0;
  cin>>n>>k;
  while(n>0){
    n/=k;
    ans++;
  }
  out(ans);
}
void solve_C() {
  int n,ans=INF;
  cin>>n;
  vi x(n);
  rep(i,n) cin>>x[i];
  sort(all(x));
  REP(a,x[0],x[n-1]+1){
    int p=0;
    rep(i,n) p+=(a-x[i])*(a-x[i]);
    chmin(ans,p);
  }
  out(ans);
}

signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  solve_B();
}