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.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.HashMap;
import java.util.Scanner;
import java.util.HashSet;
import java.util.Iterator;

class Main {

    static final int MOD = 1000000007;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long N  = sc.nextLong();
        ArrayList<Long> prime = divisorX(N);
        long min = N*N;
        long q1 = 0;
        long q2 = 0;
        for(int i=1;i<prime.size();i+=2){
            if(prime.get(i-1)+prime.get(i) < min){
                min = prime.get(i-1)+prime.get(i);
                q1 = prime.get(i-1);
                q2 = prime.get(i);
            }
        }
        // System.out.println(q1);
    
        // System.out.println(q2);
        System.out.println(min-2);
    }

    public static final ArrayList<Long> divisorX(final long n) {
		final ArrayList<Long> list = new ArrayList<Long>();

		for (long i = 1; i * i <= n; i++) {
			if (n % i == 0) {
				list.add(i);
				if (i != n / i) {
					list.add(n / i);
				}
			}
		}
		return list;
	}
}