Problem:
Score : 300 points

Problem Statement

Takahashi, Aoki and Snuke love cookies. They have A, B and C cookies, respectively. Now, they will exchange those cookies by repeating the action below:

Each person simultaneously divides his cookies in half and gives one half to each of the other two persons.

This action will be repeated until there is a person with odd number of cookies in hand.

How many times will they repeat this action?
Note that the answer may not be finite.

Constraints

1 ≤ A,B,C ≤ 10^9

Input

Input is given from Standard Input in the following format:

A B C

Output

Print the number of times the action will be performed by the three people, if this number is finite.
If it is infinite, print -1 instead.

Sample Input 1

4 12 20

Sample Output 1

3

Initially, Takahashi, Aoki and Snuke have 4, 12 and 20 cookies. Then,

After the first action, they have 16, 12 and 8.

After the second action, they have 10, 12 and 14.

After the third action, they have 13, 12 and 11.

Now, Takahashi and Snuke have odd number of cookies, and therefore the answer is 3.

Sample Input 2

14 14 14

Sample Output 2

-1

Sample Input 3

454 414 444

Sample Output 3

1

Solution in Go (1.6) (judge verdict: Accepted)

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 pow(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
}

//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 A, B, C int

func main() {
	sc.Split(bufio.ScanWords)
	A, B, C = nextInt(), nextInt(), nextInt()

	pfsA := PrimeFactors(A + B)
	pfsB := PrimeFactors(B + C)
	pfsC := PrimeFactors(C + A)

	if A%2 == 1 || B%2 == 1 || C%2 == 1 {
		fmt.Println(0)
		return
	}

	if A == B && B == C {
		fmt.Println(-1)
		return
	}

	if A%2 == 1 || B%2 == 1 || C%2 == 1 {
		fmt.Println(0)
		return
	}

	var a2, b2, c2 int
	for _, p := range pfsA {
		if p == 2 {
			a2++
		}
	}

	for _, p := range pfsB {
		if p == 2 {
			b2++
		}
	}

	for _, p := range pfsC {
		if p == 2 {
			c2++
		}
	}

	fmt.Println(minInt(a2, minInt(b2, c2)))
}