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;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep2(i, s, n) for (int i = s; i < (int)(n); i++)
#define rep3(i, s, n) for (int i = s; i > (int)(n); i--)
#define all(obj) obj.begin(), obj.end()
#define db(x) cerr << #x << ":" << x << " "
#define dbl(x) cerr << #x << ":" << x << "\n"
#define dbv(vec) cerr << #vec << ":"; for (auto e : vec) cerr << e << " "; cout << "\n"
#define dbvv(vv) cerr << #vv << ":\n"; for (auto vec : vv) { for (auto e : vec) cerr << e << " "; cerr << endl; }
#define YN(f) cout << (f ? "YES" : "NO") << endl
#define Yn(f) cout << (f ? "Yes" : "No") << endl
#define yn(f) cout << (f ? "yes" : "no") << endl
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using pii = pair<int, int>;

int main() {
	long long N; cin >> N;
	ll m = (int)sqrt(N) + 1;
	assert(m * m >= N);
	while (N % m) {
		m--;
	}
	ll ans = m + N / m - 2;
	cout << ans << endl;
}