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<algorithm>//sort(all(変数),greater<型名>())　で降順に
#include<cmath>//切り上げceil(値)
#include<deque>
#include<iomanip>//setprecision(数字)
#include<iostream>
#include<list>
#include<map>
#include<numeric> //xとyの最大公約数は__gcd(x,y),xとyの最小公倍数は(x*y)/__gcd
#include<queue>//priority_queue<型名> 変数：優先度付きキュー　最大値の探索が早い
#include<set>
#include<stack>
#include<string>
#include<vector>

#include<bits/stdc++.h>
#include<cstdio>
#include<cstdlib>
#include<limits>
#include<tuple>
#include<utility>
#define rep0(i,n)  for(ll i=0; i<n; i++)
#define rrep0(i,n) for(ll i=n-1; 0<=i; i--)
#define rep1(i,n)  for(ll i=1; i<=n; i++)
#define rrep1(i,n) for(ll i=n; 1<=i; i--)
#define vll vector<ll>
#define vi vector<int>
#define all(x) (x).begin(),(x).end()
#define mod 1000000007
#define INF 1e18
using ll = long long;
using namespace std;

int main(){
  ll n; cin >> n;
  ll ans = INF;
  rep1(i,n){
    if(n % i == 0){
      ll a = i-1+n/i-1;
      if(ans < a){
        return 0;
      }else{
        ans = a;
      }
    }
  }
  cout << ans;
}