ORIGIN

CodeForces-1293B JOE is on TV!

2020-01-19

ACM Simulation CodeForces 2 mins376 words

B. JOE is on TV!
time limit per test: 1 second
memory limit per test: 256 megabytes
input: standard input
output:standard output

3R2 - Standby for Action

Our dear Cafe’s owner, JOE Miller, will soon take part in a new game TV-show “1 vs. n”!

The game goes in rounds, where in each round the host asks JOE and his opponents a common question. All participants failing to answer are eliminated. The show ends when only JOE remains (we assume that JOE never answers a question wrong!).

For each question JOE answers, if there are $s (s>0)$ opponents remaining and $t (0≤t≤s)$ of them make a mistake on it, JOE receives $\frac{t}{s}$ dollars, and consequently there will be $s−t$ opponents left for the next question.

JOE wonders what is the maximum possible reward he can receive in the best possible scenario. Yet he has little time before show starts, so can you help him answering it instead?

Input

The first and single line contains a single integer $n (1≤n≤10^5)$, denoting the number of JOE’s opponents in the show.

Output

Print a number denoting the maximum prize (in dollars) JOE could have.

Your answer will be considered correct if it’s absolute or relative error won’t exceed $10^{−4}$. In other words, if your answer is aa and the jury answer is bb, then it must hold that $\frac{|a−b|}{max(1,b)}≤10^{−4}$.

Examples

input

output

1	1.000000000000

input

output

1	1.500000000000

Note

In the second example, the best scenario would be: one contestant fails at the first question, the other fails at the next one. The total reward will be $\frac{1}{2}+\frac{1}{1}=1.5$ dollars.

Analysis

For the biggest, only if there is one person who answer the wrong can he go through the most questions. So the longest time he can go through, the largest sum of money he can get.

Code

#include<bits/stdc++.h>
 
using namespace std;
 
int main() {
    int n;
    cin >> n;
    double ans = 0;
    while(n) {
    	ans += 1.0 / n;
    	n --;
	}
	printf("%.12f\n", ans);
    return 0;
}