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)

#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define RITR(x,c) for(__typeof(c.rbegin()) x=c.rbegin();x!=c.rend();x++)
#define setp(n) fixed << setprecision(n)

#define lf double
#define ll long long
#define vll vector<ll>
#define vi vector<int>
#define pll pair<ll,ll>
#define pi pair<int,int>

#define all(a) (a.begin()),(a.end())
#define rall(a) (a.rbegin()),(a.rend())
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define ins insert

using namespace std;

int main(void)
{
	cin.tie(0);
	ios::sync_with_stdio(false);
	ll n; cin>>n;
	ll ans=0;
	for(int i=(int)(sqrt(n)); i>=1; i--){
		if (n%i==0){
			ans += i-1;
			ans += n/i-1;
			break;
		}
	}
	cout << ans << "\n";
	return 0;
}