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)

#C問題
#考え方：平方根に近いところを探す。
#平方根を切り上げた値からスタート。割り切れた時点で終了。
#TLEになった。原因として考えられるのは何か？ものすごい大きな数の時にうまくいかない？
#素数の時に厳しい。
import math
def make_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
import numpy as np


def getNearestValue(list, num):
    """
    概要: リストからある値に最も近い値を返却する関数
    @param list: データ配列
    @param num: 対象値
    @return 対象値に最も近い値
    """

    # リスト要素と対象値の差分を計算し最小値のインデックスを取得
    idx = np.abs(np.asarray(list) - num).argmin()
    return list[idx]

N = int(input())
lis_1 = make_divisors(N)
A = getNearestValue(lis_1,math.sqrt(N))
B = int(N/A)

print(A+B-2)