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 Edge = pair<int, int>;
using Graph = vector<vector<Edge>>;
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define SORT(v) sort((v).begin(), (v).end())
#define RSORT(v) sort((v).rbegin(), (v).rend())
const ll MOD = 1000000007;
const ll nmax = 8;
const ll INF = 1e9;
bool graph[nmax][nmax];
vector<vector<ll>> dist = vector<vector<ll>>(nmax, vector<ll>(nmax, INF));
void warshall_floyd(ll n)
{
    for (size_t i = 0; i < n; i++)
    {
        for (size_t j = 0; j < n; j++)
        {
            for (size_t k = 0; k < n; k++)
            {
                dist[j][k] = min(dist[j][k], dist[j][i] + dist[i][k]);
            }
        }
    }
}

class UnionFind
{
public:
    vector<ll> Parent;

    UnionFind(ll N)
    {
        Parent = vector<ll>(N, -1);
    }
    ll find(ll A)
    {
        if (Parent[A] < 0)
            return A;
        return Parent[A] = find(Parent[A]);
    }

    ll size(ll A)
    {
        return -Parent[find(A)];
    }

    bool Union(ll A, ll B)
    {
        A = find(A);
        B = find(B);
        if (A == B)
        {
            return false;
        }
        if (size(A) < size(B))
            swap(A, B);

        Parent[A] += Parent[B];
        Parent[B] = A;

        return true;
    }
};

ll gcd(ll a, ll b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
    ll g = gcd(a, b);
    return a / g * b;
}

ll mulMod(ll a, ll b)
{
    return (((a % MOD) * (b % MOD)) % MOD);
}

ll powMod(ll a, ll p)
{
    if (p == 0)
    {
        return 1;
    }
    else if (p % 2 == 0)
    {
        ll half = powMod(a, p / 2);
        return mulMod(half, half);
    }
    else
    {
        return mulMod(powMod(a, p - 1), a);
    }
}

ll ceil(ll a, ll b)
{
    return (a + b - 1) / b;
}

void solve(long long n)
{

    ll num = n;
    ll ans = abs(n - 1);

    ll tmp = 1;
    for (ll i = 1; i * i <= n; i++)
    {
        if (num % i == 0)
        {
            ans = min(ans, abs(i - 1 + (num / i) - 1));
        }
    }

    cout << ans << endl;
}

int main()
{
    long long N;
    scanf("%lld", &N);
    solve(N);
    return 0;
}