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;
using ll = long long;
using vi = vector<int>;
using vc = vector<char>;
using vb = vector<bool>;
using vd = vector<double>;
using vs = vector<string>;
using vpii = vector<pair<int, int>>;
using vvi = vector<vector<int>>;
using vvc = vector<vector<char>>;
using vvs = vector<vector<string>>;
using pii = pair<int, int>;
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define reps(i,n) for(int i = 1; i <= (int)(n); i++)
#define rrep(i,n) for(int i = (int)(n) - 1; i >= 0; i--)
#define rreps(i,n) for(int i = (int)(n); i > 0; i--)
#define FOR(i,a,b) for(int i = (int)(a); i <= (int)(b); i++)
#define RFOR(i,a,b) for(int i = (int)(a); i >= (int)(b); i--)
#define all(x) (x).begin(), (x).end()
#define MAX(x) *max_element(all(x))
#define MIN(x) *min_element(all(x))
template <class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template <class T> bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
const int INF = INT_MAX;
const int MOD = 1000000007;

int main() {
    ll N;
    cin >> N;
    vector<pair<long, long>> p;
    for (long i = 1; i*i <= N; i++) {
        if (N % i == 0) {
            p.push_back(make_pair(i, N/i));
        }
    }
    ll ans = __LONG_LONG_MAX__;
    rep(i, p.size()) {
        ll d = (p[i].first - 1) + (p[i].second - 1);
        chmin(ans, d);
    }
    cout << ans << endl;
}