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: Accepted)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author
 */

#include <iostream>
#include <fstream>

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using pii = pair<int,int>;
using pll = pair<ll,ll>;
using vi = vector<int>;
using vl = vector<ll>;

using _loop_int = int;
#define REP(i,n) for(_loop_int i=0; i<(_loop_int)(n); i++)
#define FOR(i,a,b) for(_loop_int i=(_loop_int)(a); i<(_loop_int)(b); i++)
#define FORR(i,a,b) for(_loop_int i=(_loop_int)(b)-1; i>=(_loop_int)(a); i--)

#define CHMIN(a,b) (a)=min((a),(b))
#define CHMAX(a,b) (a)=max((a),(b))
#define ALL(v) (v).begin(),(v).end()

#define DEBUG(x) cerr<<#x<<": "<<(x)<<endl
#define DEBUG_VEC(v) cerr<<#v<<": ";REP(__i,(v).size())cerr<<((v)[__i])<<", ";cerr<<endl

const ll MOD = 1000000007ll;

class CWalkOnMultiplicationTable {
public:
    void solve(std::istream& in, std::ostream& out) {
        ll n;
        in >> n;
        ll ans = n-1;
        FOR(x,1,1000025){
            if(n%x != 0)continue;
            ll y = n / x;
            CHMIN(ans, x-1 + y-1);
        }
        out << ans << endl;
    }
};


int main() {
    std::ios_base::sync_with_stdio(false);
    CWalkOnMultiplicationTable solver;
    std::istream& in(std::cin);
    std::ostream& out(std::cout);
    in.tie(0);
    out << std::fixed << std::setprecision(10);
    solver.solve(in, out);
    return 0;
}