[3R2 - Standby for Action](https://www.youtube.com/watch?v=P2ZVC9aoiKo)

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 \displaystylet/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 a and the jury answer is b, then it must hold that (|a - b|)/(max(1, b)) ≤ 10^{-4}.

Examples

Input


1


Output


1.000000000000


Input


2


Output


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 \displaystyle 1/2 + 1/1 = 1.5 dollars.
# ===============================================================================================
# importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import *
from copy import *
from collections import deque
from collections import Counter as counter  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb  # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
# If the element is already present in the list,

# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect

# If the element is already present in the list,
# the right most position where element has to be inserted is returned

# ==============================================================================================
# fast I/O region

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None


    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

# inp = lambda: sys.stdin.readline().rstrip("\r\n")

# ===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType


def iterative(f, stack=[]):
    def wrapped_func(*args, **kwargs):
        if stack: return f(*args, **kwargs)
        to = f(*args, **kwargs)
        while True:
            if type(to) is GeneratorType:
                stack.append(to)
                to = next(to)
                continue
            stack.pop()
            if not stack: break
            to = stack[-1].send(to)
        return to

    return wrapped_func


#### END ITERATE RECURSION ####

# ===============================================================================================
# some shortcuts

mod = 1000000007


def inp(): return sys.stdin.readline().rstrip("\r\n")  # for fast input

def out(var): sys.stdout.write(str(var))  # for fast output, always take string

def lis(): return list(map(int, inp().split()))

def stringlis(): return list(map(str, inp().split()))

def sep(): return map(int, inp().split())

def strsep(): return map(str, inp().split())


def zerolist(n): return [0] * n

def nextline(): out("\n")  # as stdout.write always print sring.

def testcase(t):
    for p in range(t):
        solve()

def printlist(a):
    for p in range(0, len(a)):
        out(str(a[p]) + ' ')

def lcm(a, b): return (a * b) // gcd(a, b)

def power(x, y, p):
    res = 1  # Initialize result
    x = x % p  # Update x if it is more , than or equal to p
    if (x == 0):
        return 0
    while (y > 0):
        if ((y & 1) == 1):  # If y is odd, multiply, x with result
            res = (res * x) % p

        y = y >> 1  # y = y/2
        x = (x * x) % p
    return res

def ncr(n, r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))

def isPrime(n):
    if (n <= 1): return False
    if (n <= 3): return True
    if (n % 2 == 0 or n % 3 == 0): return False
    i = 5
    while (i * i <= n):
        if (n % i == 0 or n % (i + 2) == 0):
            return False
        i = i + 6
    return True


def solve():
    n=int(input())
    ans=0
    for i in range(1,n+1):
        ans+=(1/i)
    print(ans)
solve()