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 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <math.h>

int main(void){
    __int64_t a=0;
    scanf("%ld",&a);
    //printf("%ld",a);

//素因数分解をしてfac[]に格納,kakunousuu
    int s = floor(sqrt(a));
    int r=0;
    int factor[100];
    int x,y=0;
    //平方数まで見れば十分
    int kakunousuu=0;
    //格納変数
    int i=0;
    for(i=2;i<=s;i++){
        
        if(a%i==0){
            //0で割れるならば，割れなくなるまで割って指数カウンタをカウントアップ
            r=0; //指数カウンタ
            do{
                a=a/i;
                r++;
                factor[kakunousuu]=i;
                kakunousuu++;
        //        printf("kakunousuu=%d,i=%d\n",kakunousuu,i);

            }while(a%i==0);
            
  //          printf("i=%dで%d回割れました\n",i,r);


        }
        
    }
    
 //   printf("factor[1]=%d\n",factor[1]);
    
    
    if(kakunousuu==0){
        printf("%ld",a+1-2);
    }else{
        for (i = 0; i < kakunousuu; i++) {
            if(i%2==1){
                x=x+factor[i];
            }else{
                y=y+factor[i];
            }
            
        }
        
  //      printf("X=%d,Y=%d\n",x,y);
        printf("%d",x+y-2);
    }
    
//fac kakunousuuの半分までを足す x 後半を足す y
    
    
//xかyどっちかが1ならx+y-1
//それ以外x+y-2
//をprint
    




}