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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
public class Main {
   public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);

      long n = sc.nextLong();

      if(n<100000000){

      long m = (long)Math.sqrt(n);

      long a = m;
      long b = m;


      for (long i = 0; i < n; i++) {
         a = a + 1;
         if (n % a == 0) {
            break;
         }
      }

      for (long i = 0; i < n; i++) {
         b = b + 1;
         if (n % b == 0) {
            break;
         }
      }

      if(a + n / a < b + n / b){
         System.out.println(a + n / a - 2);
      }
      else{
         System.out.println(b + n / b - 2);
      }
   }

   else{
      if(n % 2 == 0){
         n = n / 2;

      long m = (long)Math.sqrt(n);

      long a = m;
      long b = m;


      for (long i = 0; i < n; i++) {
         a = a + 1;
         if (n % a == 0) {
            break;
         }

      }

      if(a * 2 + n / a < a + n * 2 / a){
         System.out.println(a * 2 + n / a - 2);
      }
      else{
         System.out.println(a + n * 2 / a - 2);
      }
   }
   else{
      
      long m = (long)Math.sqrt(n);

      long a = m;
      if(a%2==0){
         a= a + 1;
      }


      for (long i = 1; i < n; i++) {
         a = a - 2;
         if (n % a == 0) {
            break;
         }
      }
         System.out.println(a + n / a - 2);

   } 


   }

   }
}