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 Perl (v5.18.2) (judge verdict: Runtime Error)

use Math::Prime::Util ':all';
use Math::BigInt;

chomp($n = <>);
$n = Math::BigInt->new($n);

if(is_prob_prime($n) != 0){
	printf ($n - 1) . "\n";
	exit;
}

@table     = (1,2,2,4);
$w_divisor = 0;
$w_power   = 0;
$divisor   = 2;
$index     = 0;

@factor = ();
while ($n > $divisor){
	$surplus = $n % $divisor;
	if ($surplus == 0){
		$n /= $divisor;
		if ($w_divisor == 0){
			$w_divisor = $divisor;
			$w_power = 1;
		}elsif ($divisor == $w_divisor){
			$w_power++;
		}elsif ($divisor != $w_divisor){
			if ($w_power == 1){
				push(@factor, $w_divisor);
			}else{
				for (1 .. $w_power){
					push(@factor, $w_divisor);
				}
			}
			$w_power = 1;
			$w_divisor = $divisor;
		}
	}else{
		$divisor += $table[$index];
		if ($index == 3){
			$index--;
		}else{
			$index++;
		}
	}
}
if ($w_divisor == 0){
	push(@factor, $n);
}elsif ($w_power == 1){
	push(@factor, $w_divisor);
	push(@factor, $n);
}elsif ($w_divisor == $n){
	$w_power++;
	for (1 .. $w_power){
		push(@factor, $w_divisor);
	}
}else{
	for (1 .. $w_power){
		push(@factor, $w_divisor);
	}
	push(@factor, $n);
}

if(scalar(@factor) == 2){
	printf (($factor[0] - 1) + ($factor[1] - 1)) . "\n";
}else{
	($a, $b) = (1, 1);
	for $i (0 .. $#factor){
		if($i % 2 == 0){
			$a *= $factor[$i];
		}else{
			$b *= $factor[$i];
		}
	}
	printf (($a - 1) + ($b - 1)) . "\n";
}