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: Wrong Answer)

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
 bool isPrime(long long n);
int main()
{
    long long n,sum=0;
    cin>>n;
if(isPrime(n))
    sum= n-1;
else if(n%2==0&&n%3==0) sum=min(1+((n/2)-1),2+((n/3)-1));
else if(n%2==0&&n%4==0) sum=min(1+((n/2)-1),3+((n/4)-1));
else if(n%2==0&&n%5==0) sum=min(1+((n/2)-1),4+((n/5)-1));

else if(n%3==0&&n%4==0) sum=min(2+((n/3)-1),3+((n/4)-1));
else if(n%3==0&&n%5==0) sum=min(2+((n/3)-1),4+((n/5)-1));

else if(n%4==0&&n%5==0) sum=min(3+((n/4)-1),4+((n/5)-1));

else if(n%2==0) sum=1+((n/2)-1);
else if(n%3==0) sum=2+((n/3)-1);
else if(n%4==0) sum=3+((n/4)-1);
else if(n%5==0) sum=4+((n/5)-1);


    cout << sum ;
    return 0;
}

 bool isPrime(long long n){
for(long long i=2;i<n/2;i++){
    if (n%i==0)return false;
    if(i==10000000) break;
}
return true;
}