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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
 
#define BIG 2000000007
 
#define MOD 1000000007
typedef unsigned long long ull;
typedef   signed long long sll;

#define N_MAX 200000
#define M_MAX 200000
 
typedef struct {
	int a;
	int b;
} hw;
 
 
const hw vector8[8] = {
	{-1, -1},
	{-1,  0},
	{-1, +1},
	{ 0, -1},
	{ 0, +1},
	{+1, -1},
	{+1,  0},
	{+1, +1}
};
 
ull n, m;
ull h, w;
ull k;
ull va, vb, vc, vd, ve, vf;
// ull a[N_MAX];
// sll a[N_MAX];
// ull b[M_MAX];
// ull dp[N_MAX][M_MAX + 1];
// char s[N_MAX + 1];
// char t[N_MAX + 1];
// char s[N_MAX][M_MAX + 1];
hw arr[N_MAX];
// hw brr[M_MAX];
// ull digitdp[102][   2][    2];
//          pos  less  carry
 
void swap_adj(ull *a, ull *b){
	ull tmp = *b;
	*b = *a;
	*a = tmp;
	return;
}
 
ull divide(ull a, ull b){
	ull x = MOD - 2;
	ull ans = 1;
	while (x) {
		if (x & 1) ans = (ans * b) % MOD;
		b = (b * b) % MOD;
		x /= 2;
	}
	return (a * ans) % MOD;
}
 
int digits(ull x){
	int i = 1;
	while (x >= 10) {
		x /= 10;
		i++;
	}
	return i;
}
 
int min(ull x, ull y){
	return (x < y) ? x : y;
}
 
ull gcd(ull x, ull y){
	if (x < y) {
		return gcd(y, x);
	} else if (y == 0) {
		return x;
	} else {
		return gcd(y, x % y);
	}
}
 
ull bitpow(ull a, ull x){
	ull result = 1;
	while (x) {
		if (x & 1) {
			result *= a;
			result %= MOD;
		}
		x /= 2;
		a = (a * a) % MOD;
	}
	return result;
}
 
// int nextroute(int arr[]){
// 	int i = n - 1;
// 	int j, x;
// 	while (arr[i - 1] > arr[i]) i--;
 
// 	x = n;
// 	for (j = i; j < n; j++) {
// 		if (arr[j] < arr[i - 1]) continue;
// 		if (x == n || arr[x] > arr[j]) x = j;
// 	}
// 	arr[i - 1] ^= arr[x];
// 	arr[x] ^= arr[i - 1];
// 	arr[i - 1] ^= arr[x];
 
// 	qsort(&arr[i], n - i, sizeof(int), comp);
// 	return 0;
// }
 
int nibutan_target(ull target){
	ull maxdist = (target * (target + 1) / 2); // 時刻targetまでに到着できる距離は[-maxdist, maxdist]
	return (n <= maxdist);
}
 
int targetdig(ull x, int index /* 1-indexed */){
	// static...?
	int posmax = digits(x);
	if (posmax < index) return -1;
	while (posmax > index) {
		posmax--;
		x /= 10;
	}
	return x % 10;
}
 
int intcomp(const void *left, const void *right){
	if ((*(int*)left) < (*(int*)right)) {
		return -1;
	} else if ((*(int*)left) > (*(int*)right)) {
		return +1;
	} else {
		return 0;
	}
}
 
int ullcomp(const void *left, const void *right){
	if ((*(ull*)left) < (*(ull*)right)) {
		return -1;
	} else if ((*(ull*)left) > (*(ull*)right)) {
		return +1;
	} else {
		return 0;
	}
}
 
int hwAcomp(const void *left, const void *right){
	return intcomp(&(((hw*)left)->a), &(((hw*)right)->a));
}
 
int hwBcomp(const void *left, const void *right){
	return intcomp(&(((hw*)left)->b), &(((hw*)right)->b));
}
 
int hwABcomp(const void *left, const void *right){
	int x = hwAcomp(left, right);
	if (x) return x;
	return hwBcomp(left, right);
}

ull solve(){
	int i, j;

	ull x = (va > vb ? va - vb : vb - va);
	ull y = (vc > vb ? vc - vb : vb - vc);

	if (x == 0 && y == 0) {
		puts("-1");
	} else {
		int ans = 0;
		while (x % 2 == 0 && y % 2 == 0) {
			ans++;
			x /= 2;
			y /= 2;
		}
		printf("%d\n", ans);
	}

	return 0;
}
 
int main(void){
	int i, j;

	scanf("%llu%llu%llu", &va, &vb, &vc);

	solve();
 
	return 0;
}