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)


#pragma warning(disable:4996)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <float.h>

#pragma region 定義
#define i_cin(X) scanf("%d", &X)
#define i_cin2(X,Y) scanf("%d %d",&X,&Y)
#define i_cin4(X,Y,Z,W) scanf("%d %d %d %d",&X,&Y,&Z,&W)
#define i_cin3(X,Y,Z) scanf("%d %d %d",&X,&Y,&Z)
#define l_cin(X) scanf("%ld",&X)
#define f_cin(X) scanf("%f",&X)
#define l_cin2(X,Y) scanf("%ld %ld",&X,&Y)
#define s_cin(X) scanf("%s",&X)
#define c_cin(X) scanf("%c",&X);
#define ll_cin(X) scanf("%lld", &X)
#define ull_cin(X) scanf("%llu", &X)
#define lf_cin(X) scanf("%lf", &X)

#define  i_cout(X) printf("%d\n",X)
#define  f_cout(X) printf("%g\n",X)
#define  i_cout2(X,Y) printf("%d %d\n",X,Y)
#define  l_cout(X) printf("%ld\n",X)
#define  s_cout(X) printf("%s\n",X)
#define  s_coutc(X) printf("%s",X)

#define  c_cout(X) printf("%c",X)


#define  ll_cout(X) printf("%lld\n",X)
#define  ull_cout(X) printf("%llu\n",X)



typedef long long  ll;
typedef unsigned long long  ull;

#define rept(x, s, n) for (int x = s; x < n; x++)

#define S_RTN(S)  s_cout(S);return 0


#define _itoa(A,N) sprintf(A, "%d", N);

int i_cins(int n, int* A);
int l_cins2(int n, long* A, long* B);
int s_dsort(const void* a, const void* b);
int s_asort(const void* a, const void* b);

int  _gcd(int a, int b);
int  _swp(int* a, int* b);
int _cknum(char* a, int n);
int _atoi(char* s, int len);


int s_asorts(const void* a, const void* b);

//昇順
int s_asorts(const void* a, const void* b) {
	return(strcmp((char*)a, (char*)b));
}
int s_dsort(const void* a, const void* b) {
	return(*(int*)b - *(int*)a);
}
int s_asort(const void* a, const void* b) {
	return(*(int*)a - *(int*)b);
}
int l_cins2(int n, long* a, long* b) {
	int i;
	rept(i, 0, n) {
		l_cin2(*(a + i), *(b + i));
	}
	return 0;
}
int i_cins(int n, int* a) {
	int i;
	for (i = 0; i < n; i++) {
		i_cin(*(a + i));
	}

	return 0;
}
int _gcd(int a, int b) {
	int r, tmp;
	if (a < b) { tmp = a; a = b; b = tmp; }
	r = a % b;
	while (r) {
		if (!r) break;
		a = b; b = r;
		r = a % b;
	}
	return(b);
}
// puts("Yes\n") puts("No\n")
// scanf("%d.%d") 浮動小数点入力を整数２つで得る


//#define pai  3.1415926535      
#define deg_rad(X)  (X*(M_PI/180))

long _max(long a, long b) {
	return a > b ? a : b;
}
long _min(long a, long b) {
	return a < b ? a : b;
}
int _swp(int* a, int* b)
{
	int tmp;
	tmp = *b; *b = *a; *a = tmp;
	return 0;
}

int _cknum(char* a, int n) {
	int i;
	char t = '0';

	for (i = 0; i < n; i++) {
		if (a[i] < '0' || a[i]>'9') return 1;
	}
	return 0;
}


int _atoi(char* s, int len) {
	char tmp[20];
	memcpy(tmp, s, len);
	tmp[len] = 0x00;
	return (atoi(tmp));
}

//素因数分解　pf素数が入る　pc 素数の数
ll _pfact(ll a);
static ll pf[100000000];
static ll pc;

ll _pfact(ll a) {
	long i, r;
	if (a % 2 == 0) {
		pf[pc] = 2; pc++;
		return(_pfact(a / 2));
	}
	r = sqrt(a);
	for (i = 3; i <= r; i += 2) {
		if (a % i == 0) {
			pf[pc] = i; pc++;
			return(_pfact(a / i));
		}
	}
	if (a != 1) {
		pf[pc] = a; pc++;
	}
	return 0;

}


#pragma endregion
//150-C
//169-A

//AGC-003 A 164

int main(void) {
	long t, a, s,cnt=0;

	scanf("%ld %ld %ld", &t, &a, &s);
	if (t == a || t == s || a == s)
	{puts("-1"); return 0;}

	long tt, aa;
	while (1) {
		if ((a + s) % 2 || (t + s) % 2 || (a + t) % 2)
			break;
		
		tt = (a + s) / 2;
		aa = (t + s) / 2;
		s =  (a + t) / 2;
		t = tt; a = aa;
	
		cnt++;
	}
	i_cout(cnt);
	return 0;
}



//int main(void) {
//	char a[1001];
//	int i, len, e=0,w=0,n=0,s=0;
//
//	scanf("%s", a);
//	len = strlen(a); a[len] = 0x00;
//	for (i = 0; i < len; i++) {
//		switch (a[i]) {
//		case 'N':n ++; break;
//		case 'S':s ++; break;
//		case 'E':e ++; break;
//		case 'W':w ++; break;
//		}
//	}
//	if ((e == w) && (n && s)) {
//		printf("Yes\n"); return 0;
//	}
//	if ((n == s) && (e && w)) {
//		printf("Yes\n"); return 0;
//	}
//	
//    printf("No\n");
//
//
//	return 0;
//}

//066-B

//157-C
//int main(void) {
//	int n, m;
//	int s, c;
//	int a[3] = { -1,-1,-1 };
//	int i, ans=0;
//	
//	i_cin2(n, m);
//
//	if  (m == 0 && n!=1) { i_cout(-1); return 0; } //???2つ目の条件なし
//
//	for (i = 0; i < m; i++) {
//		i_cin2(s, c);
//		if ((a[s - 1] != -1) &&   //矛盾はないか
//			(a[s - 1] != c)  ) { i_cout(-1); return 0; }
//		switch (n) {
//		case 1:if (s != 3) { i_cout(-1); return 0; } 
//			   break;
//		case 2:if (s == 1) { i_cout(-1); return 0; }
//			  if (s == 2 && c == 0)
//			  {i_cout(-1); return 0;}
//			  break;
//		case 3:
//			if (s == 1 && c == 0)
//			{i_cout(-1); return 0;}
//			break;
//		}
//		a[s - 1] = c;
//	}
//
//	for (i = 0; i < 3; i++) {
//		if (a[i] == -1) a[i] = 0;
//	}
//	ans = a[0] * 100 + a[1] * 10 + a[2];
//
//	i_cout(ans);
//	return 0;
//}



 //ABC-07 D 幅優先探索
//#define Q_SIZE 100000
//static struct _queue {
//	char *d[Q_SIZE];
//	int h;
//	int t;
//} q;
//
//int  pushq(struct _queue* q, char *item);
//int  pushq(struct _queue* q, char *item) {
//	if (q->t == Q_SIZE) return -1;
//	q->d[q->t] = item;
//	q->t++;
//	return 0;
//}
//
//char* popq(struct _queue* q);
//char* popq(struct _queue* q) {
//	char* tmp;
//	if (q->h == q->t) return -1;
//	tmp = q->d[q->h];
//	q->h--;
//	return tmp;
//}
//
//static struct _p {
//	int x;
//	int y;
//	int cnt;
//} q;
//static char a[51][51];
//static struct _p p[50];
//static int    pcnt;
//
//int _enq(struct _p *p, int x, int y, int cnt);
//int _enq(struct _p *p,int x, int y, int cnt) {
//	p->x = x; p->y = y; p->cnt = cnt;
//	pushq(&q,p); pcnt++;
//}
//int _deq(struct _p* p);
//int _deq(struct _p* p) {
//	p=popq(&q);
//	return(p == -1 ? p : 0);
//}
//int sqr_chk(int x, int y, int cnt) {
//	int tmp;
//	tmp = pcnt;
//	if (a[x - 1][y - 1] == '.') 
//	{
//		_enq(&p[pcnt], x - 1, y - 1, cnt); a[x - 1][y - 1] = 'X';
//	}
//	if (a[x - 1][y] == '.')
//	{
//		_enq(&p[pcnt], x - 1, y, cnt); a[x - 1][y] = 'X';
//	}
//	if (a[x+1][y] == '.') 
//	{
//		_enq(&p[pcnt], x + 1, y, cnt); a[x + 1][y] = 'X';
//	}
//	if (a[x][y+1] == '.') 
//	{
//		_enq(&p[pcnt], x, y + 1, cnt); a[x][y + 1] = 'X';
//	}
//	if (pcnt == tmp) return -1;
//	else             return 0;
//}
//
//int main(void) {
//	int r, c;
//	int sy, sx, gy, gx;
//	int i = 0, j = 0, x = 0, y = 0, ans = 0;
//	char cnt = '0';
//
//	i_cin(r, c);
//	i_cin4(sx, sy, gx, gy);
//	if (sx == gx && sy == gy) {
//		i_cout(0); return 0;
//	}
//	for (i = 0; i < r; i++)
//		s_cin(a[i]);
//	sy--; sx--;
//	struct _p* pt;
//	pt = &p[cnt];
//	p->x = sx; p->y = sy; p->cnt = cnt;
//
//	int rc;
//	while (1) {
//		rc = sqr_chk(pt->x,pt->y,pt->cnt);
//		if (rc == -1) {
//			rc = _deq(pt);
//			if (rc == -1 || (pt->x == gx && pt->y == gy)) break;
//		}
//	}
//
//	if (rc == -1) i_cout(rc);
//	else i_cout(ans);
//	return 0;
//}