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: Runtime Error)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;

#define rep(i, n) REP(i, 0, n)
#define ALL(v) v.begin(), v.end()
#define MSG(a) cout << #a << " " << a << endl;
#define REP(i, x, n) for (int i = x; i < n; i++)
#define OP(m) cout << m << endl
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

const int n = 1e6 + 5;
int p[n + 1];
ll fact[n + 1];

int main()
{

    rep(i, n + 1) p[i] = 1;
    p[0] = 0, p[1] = 0;
    for (int i = 2; i <= (int)sqrt(n); i++)
    {
        if (p[i] == 1)
        {
            for (int j = 2; i * j <= n; j++)
            {
                p[i * j] = 0;
            }
        }
    }

    ll m;
    cin >> m;

    bool flag = 0;
    for (int i = 2; i <= n; i++)
        if (m % i == 0)
            flag = 1;

    if (flag == 0)
    {
        OP(m - 1);
        return 0;
    }

    rep(i, n) fact[i] = 0;

    int k = 2;
    while (m != 1)
    {
        if (p[k] == 0)
        {
            k++;
            continue;
        }
        if (m % k == 0)
        {
            m /= k;
            fact[k]++;
        }
        else
        {
            k++;
        }
    }

    ll a = 1, b = 1;

    for (ll i = n; i >= 2; i--)
    {
        if (fact[i] >= 1)
        {
            if (b <= a)
                swap(a, b);

            ll c = ceil((double)fact[i] / 2);
            a = a * pow(i, c);
            if (fact[i] != 1)
                b = b * pow(i, (fact[i] - c));
        }
    }

    OP(a + b - 2);

    return 0;
}