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)

#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
/*#pragma GCC optimize("Ofast","unroll-loops","omit-frame-pointer","inline") //Optimization flags
#pragma GCC option("arch=native","tune=native","no-zero-upper") //Enable AVX
#pragma GCC target("avx2")  //Enable AVX*/

using namespace __gnu_pbds;
using namespace std;

#define vi vector<int>


using ll = long long;
#define ld long double
#define mp make_pair
#define what_is(x) cerr << #x << " is " << x << endl;

typedef tree<
        pair<ll, int>,
        null_type,
        less<pair<ll, int>>,
        rb_tree_tag,
        tree_order_statistics_node_update>
        ordered_set;

int p;


int mul(int a, int b) {
    return (1LL * a * b) % p;
}

int add(int a, int b) {
    int s = (a+b);
    if (s>=p) s-=p;
    return s;
}

int sub(int a, int b) {
    int s = (a+p-b);
    if (s>=p) s-=p;
    return s;
}

int po(int a, int deg)
{
    if (deg==0) return 1;
    if (deg%2==1) return mul(a, po(a, deg-1));
    int t = po(a, deg/2);
    return mul(t, t);
}

int inv(int n)
{
    return po(n, p-2);
}


vector<int> Z(vector<int> s)
{
    int n = s.size();
    vector<int> z(n);
    int L = 0, R = 0;
    for (int i = 1; i < n; i++) {
        if (i > R) {
            L = R = i;
            while (R < n && s[R-L] == s[R]) R++;
            z[i] = R-L; R--;
        } else {
            int k = i-L;
            if (z[k] < R-i+1) z[i] = z[k];
            else {
                L = i;
                while (R < n && s[R-L] == s[R]) R++;
                z[i] = R-L; R--;
            }
        }
    }
    return z;
}

mt19937 rnd(time(0));

//******CONVEX HULL TRICK******//

/*
struct Line {
    mutable ll k, m, p;
    bool operator<(const Line& o) const { return k < o.k; }
    bool operator<(ll x) const { return p < x; }
};

struct CHT : multiset<Line, less<>> {
    // (for doubles, use inf = 1/.0, div(a,b) = a/b)
    const ll inf = LLONG_MAX;
    ll div(ll a, ll b) { // floored division
        return a /b - ((a ^ b) < 0 && a % b); }
    bool isect(iterator x, iterator y) {
        if (y == end()) { x->p = inf; return false; }
        if (x->k == y->k) x->p = x->m > y->m ? inf : -inf;
        else x->p = div(y->m - x->m, x->k - y->k);
        return x->p >= y->p;
    }
    void add(ll k, ll m) {
        auto z = insert({k, m, 0}), y = z++, x = y;
        while (isect(y, z)) z = erase(z);
        if (x != begin() && isect(--x, y)) isect(x, y = erase(y));
        while ((y = x) != begin() && (--x)->p >= y->p)
        isect(x, erase(y));
    }
    ll query(ll x) {
        assert(!empty());
        auto l = *lower_bound(x);
        return l.k * x + l.m;
    }
};
*/


//******MAXFLOW******//

/*
typedef vector<int> VI;
typedef vector<VI> VVI;

const int INF = 1000000000;

struct MaxFlow {
  int N;
  VVI cap, flow;
  VI dad, Q;

  MaxFlow(int N) :
    N(N), cap(N, VI(N)), flow(N, VI(N)), dad(N), Q(N) {}

  void AddEdge(int from, int to, int cap) {
    this->cap[from][to] += cap;
  }

  int BlockingFlow(int s, int t) {
    fill(dad.begin(), dad.end(), -1);
    dad[s] = -2;

    int head = 0, tail = 0;
    Q[tail++] = s;
    while (head < tail) {
      int x = Q[head++];
      for (int i = 0; i < N; i++) {
        if (dad[i] == -1 && cap[x][i] - flow[x][i] > 0) {
          dad[i] = x;
          Q[tail++] = i;
        }
      }
    }

    if (dad[t] == -1) return 0;

    int totflow = 0;
    for (int i = 0; i < N; i++) {
      if (dad[i] == -1) continue;
      int amt = cap[i][t] - flow[i][t];
      for (int j = i; amt && j != s; j = dad[j])
        amt = min(amt, cap[dad[j]][j] - flow[dad[j]][j]);
      if (amt == 0) continue;
      flow[i][t] += amt;
      flow[t][i] -= amt;
      for (int j = i; j != s; j = dad[j]) {
        flow[dad[j]][j] += amt;
        flow[j][dad[j]] -= amt;
      }
      totflow += amt;
    }

    return totflow;
  }

  int GetMaxFlow(int source, int sink) {
    int totflow = 0;
    while (int flow = BlockingFlow(source, sink))
      totflow += flow;
    return totflow;
  }
};*/


//******DSU******//


/*
struct DSU
{
    vector<int> sz;
    vector<int> parent;

    void make_set(int v) {
        parent[v] = v;
        sz[v] = 1;
    }

    int find_set(int v) {
        if (v == parent[v])
            return v;
        return parent[v] = find_set(parent[v]);
    }

    void union_sets(int a, int b) {
        a = find_set(a);
        b = find_set(b);
        if (a != b) {
            if (sz[a] < sz[b])
                swap(a, b);
            parent[b] = a;
            sz[a] += sz[b];
        }
    }

    DSU (int n)
    {
       sz.resize(n);
        for (int i = 0; i<n; i++) make_set(i);
    }
};*/

//******MODULAR FFT******//

/*
const int root = 646;

const int root_1 = 208611436;

const int root_pw = 1<<20;


void fft (vector<int> & a, bool invert) {
    int n = (int) a.size();

    for (int i=1, j=0; i<n; ++i) {
        int bit = n >> 1;
        for (; j>=bit; bit>>=1)
            j -= bit;
        j += bit;
        if (i < j)
            swap (a[i], a[j]);
    }

    for (int len=2; len<=n; len<<=1) {
        int wlen = invert ? root_1 : root;
        for (int i=len; i<root_pw; i<<=1)
            wlen = int (wlen * 1ll * wlen % p);
        for (int i=0; i<n; i+=len) {
            int w = 1;
            for (int j=0; j<len/2; ++j) {
                int u = a[i+j],  v = int (a[i+j+len/2] * 1ll * w % p);
                a[i+j] = u+v < p ? u+v : u+v-p;
                a[i+j+len/2] = u-v >= 0 ? u-v : u-v+p;
                w = int (w * 1ll * wlen % p);
            }
        }
    }
    if (invert) {
        int nrev = po(n, p - 2);
        for (int i=0; i<n; ++i)
            a[i] = int (a[i] * 1ll * nrev % p);
    }
}
void poly_mult(vector < int >& a, vector < int > b) {
    int s = a.size() + b.size();
    int r = 1;
    while (r < s) r *= 2;
    a.resize(r);
    b.resize(r);
    fft(a, false);
    fft(b, false);
    for (int j = 0; j < r; j++) {
        a[j] = mul(a[j], b[j]);
    }
    fft(a, true);
    while (!a.empty() && (a.back() == 0)) a.pop_back();
}

*/

vector<int> suffix_array(string s)
{
    int n = s.size();

    for (int i = 0; i<n; i++)
    {

    }

}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(nullptr);


    int a, b, c;
    cin>>a>>b>>c;
    if (a==b && b == c && a%2==0) {cout<<-1; return 0;}
    int cnt = 0;
    while (a%2==0 && b%2==0 && c%2==0)
    {
        int a1 = a; int b1 = b; int c1 = c;
        a = (b1+c1)/2; b = (a1+c1)/2; c = (a1+b1)/2;
        cnt++;
    }
    cout<<cnt;



}