Problem:
Score : 300 points

Problem Statement

Takahashi is standing on a multiplication table with infinitely many rows and columns.

The square (i,j) contains the integer i \times j. Initially, Takahashi is standing at (1,1).

In one move, he can move from (i,j) to either (i+1,j) or (i,j+1).

Given an integer N, find the minimum number of moves needed to reach a square that contains N.

Constraints

2 \leq N \leq 10^{12}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the minimum number of moves needed to reach a square that contains the integer N.

Sample Input 1

10

Sample Output 1

5

(2,5) can be reached in five moves. We cannot reach a square that contains 10 in less than five moves.

Sample Input 2

50

Sample Output 2

13

(5, 10) can be reached in 13 moves.

Sample Input 3

10000000019

Sample Output 3

10000000018

Both input and output may be enormous.

Solution in Python (3.4.3) (judge verdict: Accepted)

def primelist(n):
    is_prime = [True] * (n + 1)
    is_prime[0] = False
    is_prime[1] = False
    for i in range(2, int(n**0.5) + 1):
        if not is_prime[i]:
            continue
        for j in range(i * 2, n + 1, i):
            is_prime[j] = False
    return [i for i in range(n + 1) if is_prime[i]]

def factorization(n):
    plist = primelist(int(n**0.5))
    factor = []
    for p in plist:
        k = 0
        while n%p==0:
            k+=1
            n//=p
        if k>0:
            factor.append((p,k))
        if n==1:
            break
    if n!=1:
        factor.append((n,1))

    return factor

def divisors(n):
    divisors = []
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i:
                divisors.append(n//i)

    divisors.sort()
    return divisors
  
n = int(input())

divs = divisors(n)
t = len(divs)
if t==2:
  print(divs[-1]-1)
elif t%2==0:
  print(divs[t//2-1]+divs[t//2]-2)
else:
  print(divs[t//2]*2-2)