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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <iomanip>
#include <algorithm>
#include <vector>

using namespace std;
typedef long long ll;

int main()
{
    ll n;
    cin >> n;

    ll answer = 0;

    ll last = n;

    for (ll i = 1; i < n; i++)
    {
        if (i > last || i > n / 2 || i * i > n)
        {
            break;
        }

        if (n % i == 0)
        {
            if (i < n - i)
            {
                answer = (i - 1) + (n / i - 1);
                last = n / i;
            }
            else
            {
                break;
            }
        }
    }

    cout << answer << endl;

    // bool flg = false;

    // for (int i = 1; i <= 9; i++)
    // {
    //     if (n % i == 0)
    //     {
    //         if ((n / i) >= 1 && (n / i) <= 9)
    //         {
    //             flg = true;
    //             break;
    //         }
    //     }
    // }

    // if (!flg)
    // {
    //     cout << "No" << endl;
    // }
    // else
    // {
    //     cout << "Yes" << endl;
    // }

    // for (int i = 0; i < n; i++)
    // {
    //     cin >> a[i];
    // }

    // vector<ll> a;
    // for (int i = 0; i < n; i++)
    // {
    //     int j;
    //     cin >> j;
    //     a.push_back(j);
    // }

    //sort(a,a+n); ソート
}