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++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/rope>

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define F0R(i, a) for (int i = 0; i < (a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= (a); i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define trav(a, x) for (auto& a : x)

#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound

#define sz(x) (int)x.size()
#define beg(x) x.begin()
#define en(x) x.end()
#define all(x) beg(x), en(x)
#define resz resize

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;
const ld PI = 4*atan((ld)1);

template<class T> void ckmin(T &a, T b) { a = min(a, b); }
template<class T> void ckmax(T &a, T b) { a = max(a, b); }

namespace io {
    // TYPE ID (StackOverflow)
    
    template<class T> struct like_array : is_array<T>{};
    template<class T, size_t N> struct like_array<array<T,N>> : true_type{};
    template<class T> struct like_array<vector<T>> : true_type{};
    template<class T> bool is_like_array(const T& a) { return like_array<T>::value; }

    // I/O 
    
    void setIn(string s) { freopen(s.c_str(),"r",stdin); }
    void setOut(string s) { freopen(s.c_str(),"w",stdout); }
    void setIO(string s = "") {
        ios_base::sync_with_stdio(0); cin.tie(0);
        if (sz(s)) { setIn(s+".in"), setOut(s+".out"); }
    }
    
    // INPUT 
    
    template<class T> void re(T& x) { cin >> x; }
    template<class Arg, class... Args> void re(Arg& first, Args&... rest);
    void re(double& x) { string t; re(t); x = stod(t); }
    void re(ld& x) { string t; re(t); x = stold(t); }
    
    template<class T> void re(complex<T>& x);
    template<class T1, class T2> void re(pair<T1,T2>& p);
    template<class T> void re(vector<T>& a);
    template<class T, size_t SZ> void re(array<T,SZ>& a);
    
    template<class Arg, class... Args> void re(Arg& first, Args&... rest) { re(first); re(rest...); }
    template<class T> void re(complex<T>& x) { T a,b; re(a,b); x = cd(a,b); }
    template<class T1, class T2> void re(pair<T1,T2>& p) { re(p.f,p.s); }
    template<class T> void re(vector<T>& a) { F0R(i,sz(a)) re(a[i]); }
    template<class T, size_t SZ> void re(array<T,SZ>& a) { F0R(i,SZ) re(a[i]); }
    
    // OUTPUT 
    
    template<class T1, class T2> ostream& operator<<(ostream& os, const pair<T1,T2>& a) {
        os << '{' << a.f << ", " << a.s << '}'; return os;
    }
    template<class T> ostream& printArray(ostream& os, const T& a, int SZ) {
        os << '{';
        F0R(i,SZ) {
            if (i) {
                os << ", ";
                if (is_like_array(a[i])) cout << "\n";
            }
            os << a[i];
        }
        os << '}';
        return os;
    }
    template<class T, size_t SZ> ostream& operator<<(ostream& os, const array<T,SZ>& a) {
        return printArray(os,a,SZ);
    }
    template<class T> ostream& operator<<(ostream& os, const vector<T>& a) {
        return printArray(os,a,sz(a));
    }
    template<class T> ostream& operator<<(ostream& os, const set<T>& a) {
        os << vector<T>(all(a)); return os;
    }
    template<class T1, class T2> ostream& operator<<(ostream& os, const map<T1,T2>& a) {
        os << vector<pair<T1,T2>>(all(a)); return os;
    }
    
    template<class T> void pr(const T& x) { cout << x << '\n'; }
    template<class Arg, class... Args> void pr(const Arg& first, const Args&... rest) { 
        cout << first << ' '; pr(rest...); 
    }
}

using namespace io;

int A,B,C;

int main() {
    // you should actually read the stuff at the bottom
    re(A,B,C);
    if (A == B && B == C && C == A && A % 2 == 0) {
        pr(-1);
        exit(0);
    }
    int ans = 0;
    while (A % 2 == 0 && B % 2 == 0 && C % 2 == 0) {
        int a = (A+B)/2, b = (B+C)/2, c = (C+A)/2;
        swap(A,a), swap(B,b), swap(C,c);
        ans ++;
    }
    pr(ans);
    // you should actually read the stuff at the bottom
}

/* stuff you should look for
    * int overflow, array bounds
    * special cases (n=1?), set tle
    * do smth instead of nothing and stay organized
*/