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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<stdio.h>
int main()
{
    long N;
    scanf("%ld",&N);
   	
  //
  long c[N];
  long b[N];
  long count=0;
   long i;
  long ans[N];
	for(i=0;i<N;i++){
	ans[i] = 1;
    }
	
  while (N % 2 == 0){ // 偶数の素数は2のみ
       // printf("2 *");
      //
        N /= 2;
      b[count] = N;
      c[count] = 2;
      ans[count] *= c[count];
        count++;
     ans[count] = ans[count-1];
    }
    for (i = 3; i <= N; i++){ // 他の素数で割っていく
        while (N % i == 0){
         //   printf("%d *", i);
        //
            N /= i;
            b[count] = N;
          c[count] = i;
          ans[count] *= c[count];
          count++;
          ans[count] = ans[count-1];
        }
    }
  
  //
  
  
  
  long check;
  long min = 10000000000000;
 // printf("count:%d\n",count);
  for(i=0;i<count;i++){
	if(min > (b[i]+ans[i])){
		min = b[i]+ans[i];
      	check = i;
     // printf("i:%d",i);
    }
  }
  
  long num = 0;
  num = num + b[check] -1;
  num = num + ans[check] -1;
  /* for(i=0;i<count;i++){
     printf("b:%d",b[i]);
     printf("c:%d",c[i]);
     printf("ans:%d",ans[i]);
     printf("\n");
   }*/
  
  printf("%ld",num);
    return 0;
}