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 Go (1.14.1) (judge verdict: Wrong Answer)

package main

import (
	"bufio"
	"fmt"
	"math"
	"os"
	"strconv"
)

//Util
func nextStr() string {
	sc.Scan()
	return sc.Text()
}

func nextInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}

func maxInt(a, b int) int {
	if a > b {
		return a
	} else {
		return b
	}
}

func minInt(a, b int) int {
	if a > b {
		return b
	} else {
		return a
	}
}

func absInt(a int) int {
	if a >= 0 {
		return a
	} else {
		return -a
	}
}

func absFloat64(a float64) float64 {
	if a >= 0 {
		return a
	} else {
		return -a
	}
}

func maxFloat64(a, b float64) float64 {
	if a > b {
		return a
	} else {
		return b
	}
}

func minFloat64(a, b float64) float64 {
	if a > b {
		return b
	} else {
		return a
	}
}

func str2Int(s string) int {
	n, _ := strconv.Atoi(s)
	return n
}

func reverse(s string) string {
	runes := []rune(s)
	for i, j := 0, len(runes)-1; i < j; i, j = i+1, j-1 {
		runes[i], runes[j] = runes[j], runes[i]
	}
	return string(runes)
}

func powInt(p, q int) int {
	return int(math.Pow(float64(p), float64(q)))
}

func isPrime(x int) bool {
	if x == 1 {
		return false
	}
	if x == 2 {
		return true
	}
	if x%2 == 0 {
		return false
	}

	b := true
	rootx := int(math.Sqrt(float64(x)))
	i := 3
	for i <= rootx {
		if x%i == 0 {
			b = false
			break
		}
		i += 2
	}
	return b
}

func PrimeFactors(n int) (pfs []int) {
	// Get the number of 2s that divide n
	for n%2 == 0 {
		pfs = append(pfs, 2)
		n = n / 2
	}

	// n must be odd at this point. so we can skip one element
	// (note i = i + 2)
	for i := 3; i*i <= n; i = i + 2 {
		// while i divides n, append i and divide n
		for n%i == 0 {
			pfs = append(pfs, i)
			n = n / i
		}
	}

	// This condition is to handle the case when n is a prime number
	// greater than 2
	if n > 2 {
		pfs = append(pfs, n)
	}

	return
}

func sumInts(x []int) int {

	total := 0
	for _, v := range x {
		total += v
	}

	return total
}

//Main
const (
	initialBufSize = 10000
	maxBufSize     = 1000000
	mod            = 1e9 + 7
)

var (
	sc *bufio.Scanner = func() *bufio.Scanner {
		sc := bufio.NewScanner(os.Stdin)
		buf := make([]byte, initialBufSize)
		sc.Buffer(buf, maxBufSize)
		return sc
	}()
)

var N int

func main() {
	sc.Split(bufio.ScanWords)

	N = nextInt()
	pfs := PrimeFactors(N)

	//fmt.Println(pfs)
	l := len(pfs)
	ans := powInt(10, 12)
	for i := 0; i < (1<<uint(l))/2; i++ {
		b := fmt.Sprintf("%b", i)
		cur := l - 1
		A, B := 1, 1
		for j := len(b) - 1; j >= 0; j-- {
			if string(b[j]) == "0" {
				A *= pfs[cur]
			} else {
				B *= pfs[cur]
			}
			cur--
		}
		A *= N / B
		//fmt.Println(b, A, B)
		ans = minInt(ans, A+B-2)
	}

	fmt.Println(ans)
}