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>
using namespace std;

typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
const int INF = int (1e9) + int (1e5);
const ll INFL = ll(2e18) + ll(1e10);
const ui MOD = 1E9 + 7;
const double EPS = 1e-9;
#define FOR(i,n) for (int i=0;i<(n);++i)
#define ROF(i,x) for(int i = (x) ; i >= 0 ; --i)
#define MP make_pair
#define ALL(a) (a).begin(),(a).end()
#define ODD(x) (((x)&1)==0?(0):(1))
#define sign(x) ((x > 0) - (x < 0))
std::mt19937_64  generator(std::chrono::system_clock::now().time_since_epoch().count());

inline ll powmod(ll a,ll b,ll mod) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a, ll b) { return a ? gcd(b%a, a): b; }
ll lcm(ll a, ll b) { return a / gcd(a,b) * b; }


void READ(bool _local){
    ios_base::sync_with_stdio(false); cin.tie(0);
#ifdef _DEBUG
    if (_local)
        freopen ("in.txt", "r", stdin);
#endif
}

int main() {
    READ(0);
    ll n;cin>>n;
    int x=sqrt(n);
    while(n%x!=0){
        --x;
    }
    cout << n/x + x-2;

}