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)

/*
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──────────███████████
──────────███████████
──────────██████─────
──────────█████████──
█───────███████──────
██────████████████───
███──██████████──█───
███████████████──────
███████████████──────
─█████████████───────
──███████████────────
────████████─────────
─────███──██─────────
─────██────█─────────
─────█─────█─────────
─────██────██────────
─────────────────────
*/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>

#define int long long
#define read(a) int a; cin >> a;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define reads(s) string s; cin >> s;
#define readarr(a, n) int a[(n) + 1] = {}; fori (i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; fori (i, 1, n) {fori (j, 1, m) cin >> a[i][j];}

#define print(a) cout << a << endl;
#define printarr(a, n) fori (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) fori (i, 1, n) {fori (j, 1, m) cout << a[i][j] << " "; cout << endl;} cout << endl;
#define all(v) v.begin(), v.end()
#define sz(s) (int)(s.size())
#define pb push_back
#define fi first
#define se second
#define rz resize
#define fori(i, a, b) for (int i = (a); i <= (b); i++)
#define ford(i, a, b) for (int i = (a); i >= (b); i--)
#define endl '\n'
#define setprec(x) cout << fixed << setprecision(x)
 
using namespace std;
using namespace __gnu_pbds; 
 
typedef tree<long long,null_type,less<long long>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef vector<pair <int, int>> vpi;
typedef vector < vector <int> > vvi;
 
const int INF = 4e18;
const int mod = 1e9+7; 

     
signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    //start from here

    read(x);
    int sum = 100;
    for (int i = 1;; i++)
    {
    	sum += (sum/100);
    	if(sum >= x) {cout << i; return 0;}
    }

    return 0;
}