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)

#include <bits/stdc++.h>
using namespace std;
const auto start_time = std::chrono::high_resolution_clock::now();
void timecalculater(){
    #ifndef ONLINE_JUDGE
    auto end_time = std::chrono::high_resolution_clock::now();
    std::chrono::duration<double> diff = end_time-start_time;
    cerr<<"Time Taken : "<<diff.count()<<"\n";
    #endif
}
void __print(int x) {cout << x;}
void __print(long x) {cout << x;}
void __print(long long x) {cout << x;}
void __print(unsigned x) {cout << x;}
void __print(unsigned long x) {cout << x;}
void __print(unsigned long long x) {cout << x;}
void __print(float x) {cout << x;}
void __print(double x) {cout << x;}
void __print(long double x) {cout << x;}
void __print(char x) {cout << '\'' << x << '\'';}
void __print(const char *x) {cout << '\"' << x << '\"';}
void __print(const string &x) {cout << '\"' << x << '\"';}
void __print(bool x) {cout << (x ? "true" : "false");}
template<typename T, typename V>
void __print(const pair<T, V> &x) {cout << '{'; __print(x.first); cout << ','; __print(x.second); cout << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cout << '{'; for (auto &i: x) cout << (f++ ? "," : ""), __print(i); cout << "}";}
void _print() {cout << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cout << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define dbg(x...) cout << "[" << #x << "] = ["; _print(x)
#else
#define dbg(x...)
#endif
#define rep(i, n) for(int i = 0; i < (n); ++i)
#define repA(i, a, n) for(int  i = a; i <= (n); ++i)
#define all(x) x.begin(), x.end()
#define sz(x) (int) (x).size()
#define fill(a,v) memset(a, v, sizeof(a))
#define fst first
#define snd second
#define mp make_pair
#define pb push_back
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<long long int> vll;
typedef vector<pii> vii;
ll p=1e9+7;
const int N=2e5+5;
int main() {
   
    cin.sync_with_stdio(0); cin.tie(0);
    cin.exceptions(cin.failbit);
    int T=1;
    // cin>>T;
    while(T--){
        ll x;cin>>x;
        ll ans=1;
        double a,b;
        a=101;
        b=100;
        double u=a/b;
        ll s=100;
        while(1){
            ll val=s*u;
            // u=u*101;
            // u=u/100;
            s=val;
            if(val>=x){
                cout<<ans;return 0;
            }
            ans++;
        }
    }   
    timecalculater();
    return 0;   
}