Problem:
Score : 200 points

Problem Statement

Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank.

The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.)

Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?

Constraints

101 \le X \le 10^{18}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X

Output

Print the number of years it takes for Takahashi's balance to reach X yen or above for the first time.

Sample Input 1

103

Sample Output 1

3

The balance after one year is 101 yen.

The balance after two years is 102 yen.

The balance after three years is 103 yen.

Thus, it takes three years for the balance to reach 103 yen or above.

Sample Input 2

1000000000000000000

Sample Output 2

3760

Sample Input 3

1333333333

Sample Output 3

1706

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

// https://ei1333.github.io/luzhiled/
// http://beet-aizu.hatenablog.com/entry/2017/01/04/113233
// http://www.prefield.com/algorithm/
#include <bits/stdc++.h>
constexpr char bn = '\n';
using namespace std;
using ll = long long;
using ld = long double;
using vl = vector<ll>;
using vd = vector<ld>;
using vs = vector<string>;
using vb = vector<bool>;
using vvl = vector<vector<ll>>;
using vvd = vector<vector<ld>>;
using vvs = vector<vector<string>>;
using vvb = vector<vector<bool>>;
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
template<class T> using vc = vector<T>;
template<class T> using vvc = vector<vector<T>>;
using Graph = vector<vector<ll>>;
const ll INF = 1LL << 60;
void init() {cin.tie(0);ios::sync_with_stdio(false);cout << fixed << setprecision(15);}
const ll tl= 1000000007;
#define REP(i, n) for (ll i = 0; i < n; i++)
#define REREP(i, n) for (ll i = n; i >= 0; i--)
#define FOR(i, a, n) for (ll i = a; i < n; i++)
#define REFOR(i, n, a) for (ll i = n; i >= a; i--)
#define CLR(mat,f) memset(mat, f, sizeof(mat))
#define IN(a, b, x) (a<=x&&x<=b)
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() ) //被り削除
#define debug cout << "line : " << __LINE__ << " debug" << endl;
#define ini(...) int __VA_ARGS__; in(__VA_ARGS__)
#define inl(...) long long __VA_ARGS__; in(__VA_ARGS__)
#define ind(...) long double __VA_ARGS__; in(__VA_ARGS__)
#define ins(...) string __VA_ARGS__; in(__VA_ARGS__)
#define inc(...) char __VA_ARGS__; in(__VA_ARGS__)
void in(){} template <typename T,class... U> void in(T &t,U &...u){ cin >> t; in(u...);}
void out(){cout << endl;} template <typename T,class... U> void out(const T &t,const U &...u){ cout << t; if(sizeof...(u)) cout << " "; out(u...);}
template<typename T>void die(T x){out(x); exit(0);}
#define in1(A) REP(i,A.size()) in(A[i]);
#define in2(A,B) REP(i,A.size()) in(A[i],B[i]);
#define in3(s,t,u) REP(i,sz(s)){in(s[i] , t[i] , u[i]);}
#define in4(s,t,u,v) REP(i,sz(s)){in(s[i] , t[i] , u[i] , v[i]);}
#define each(x,v) for(auto& x : v)
#define all(v) (v).begin(),(v).end()
#define sz(v) ((int)(v).size())
struct Point{ ld x,y; };
ld dist(Point a, Point b){return sqrt(abs(a.x-b.x)*abs(a.x-b.x)+abs(a.y-b.y)*abs(a.y-b.y));} // ２点間の距離 
ll gcd(ll a, ll b) { return b != 0 ? gcd(b, a % b) : a; }
ll lcm(ll a,ll b){ return a / gcd(a,b) * b;} 
ll modpow(ll a, ll n, ll mod) {
  ll res = 1;
   while (n > 0) {
      if (n & 1) res = res * a % mod;
      a = a * a % mod;
      n >>= 1;
  }
  return res;
}
ll fact(ll n){ if(n < 2) return 1; return (n * fact(n-1))%tl; } //階乗
inline ll updiv(ld a,ld b){ return (a + b - 1) / b; } //切り上げ
template<typename T,typename U>ll ceil(T a,U b){return (a + b - 1) / b;}
template <class A, class B> inline bool chmax(A &a, const B &b) { return b > a && (a = b, true); }
template <class A, class B> inline bool chmin(A &a, const B &b) { return b < a && (a = b, true); }
#define YES(n) ((n) ? "YES" : "NO"  )
#define Yes(n) ((n) ? "Yes" : "No"  )
#define yes(n) ((n) ? "yes" : "no"  )
//-------------------------------------------------------------------------------------------------
int main(){
    init();
    ll ans= 0;
    inl(n);
    ll m=100;
    // vl A(n); in1(A);
    while(n>m){
        m*=1.01;
        ans++;
    }
    out(ans);
}