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<bits/stdc++.h>
using namespace std;

#define ll long long int
#define watch(x) cout << (#x) << " is " << (x) << endl
#define f(i,x,n)  for(int i=x;i<n;i++)
#define FASTIO cin.tie(0); cout.tie(0);
#define eb(x) emplace_back(x)
#define mp(a,b) make_pair(a,b)
#define sz(a) int((a).size())
#define mod 1000000007

//MIN-MAX
#define max3(a, b, c) max(a, max(b, c))
#define min3(a, b, c) min(a, min(b, c))

//LCM
int lcm(int a, int b) {  return a * (b / __gcd(a, b)); }

//For sort by second
bool cmp(const pair<int,int> &a)
{ return a.first<a.second; }

//Tree
/* const int MAXN = 1e5 + 5;
vector<ll> adj[MAXN];*/

//ModPower
ll modpower(ll a,ll b,ll c)
{
	ll res=1;
	while(b)
	{
		if(b&1LL)
			res=(res*a)%c;
		a=(a*a)%c;
		b>>=1;
	}
	return res;
}

template<typename T> void print(T n)
{
   cout<<n;

    putchar('\n');
}
ll num(ll n)
{

    for (ll i=1; i<=sqrt(n); i++)
    {
        if (n%i==0)
        {
           ll x=n/i;
           if(x*x>n)
            return x;
        }
    }

}
int main()
{
	FASTIO;

	    ll n;
	     cin>>n;

    ll ans= num(n);

    cout<<ans+(n/ans)-2<<"\n";

	return 0;
}