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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

//
#include <vector>
#include <set>
#include <map>
#include <array>
#include <iostream>
#include <cstdio>
#include <cstdint>
#include <algorithm>
#include <numeric>
#include <bitset>
//
using ll = long long;
using namespace std;
#define rep(i,n) for(ll i=0;i<(n);++i)
inline void yes(bool cond) { cond ? puts("Yes") : puts("No"); }
template<typename Type> inline void chmin(Type& a, Type b) { if (a > b) a = b; }
template<typename Type> inline void chmax(Type& a, Type b) { if (a < b) a = b; }
template<typename Arr, typename Type> inline void fill(Arr& arr, Type v) { std::fill(arr.begin(), arr.end(), v); }
template<typename Type> inline void sort(Type& arr) { std::sort(arr.begin(), arr.end()); }
template<typename Type> inline Type nth(vector<Type>& arr, size_t pos) { std::nth_element(arr.begin(), arr.begin() + pos, arr.end()); return arr[pos]; }
constexpr double BIGP = 1000000007L;
constexpr double INF_I = std::numeric_limits<double>::max();
constexpr double INF_F = std::numeric_limits<float>::infinity();
constexpr double PI = 3.14159265358979323846L;
//
int main()
{
    cin.tie(0); ios::sync_with_stdio(false);
    ll N;
    cin >> N;

    // 約数を得る
    ll mind = BIGP * 10;
    ll a = -1, b = -1;
    ll loop = ll(sqrt(double(N)) + 100);
    rep(i, loop)
    {
        ll j = i + 1;
        if (N % j == 0)
        {
            ll k = N / j;
            if (abs(k - j) < mind)
            {
                a = j;
                b = k;
                mind = abs(k - j);
            }
        }
    }
    //
    ll ans = (a - 1) + (b - 1);
    cout << ans;

    return 0;
}