Problem: Arseny likes to organize parties and invite people to it. However, not only friends come to his parties, but friends of his friends, friends of friends of his friends and so on. That's why some of Arseny's guests can be unknown to him. He decided to fix this issue using the following procedure.

At each step he selects one of his guests A, who pairwise introduces all of his friends to each other. After this action any two friends of A become friends. This process is run until all pairs of guests are friends.

Arseny doesn't want to spend much time doing it, so he wants to finish this process using the minimum number of steps. Help Arseny to do it.

Input

The first line contains two integers n and m (1 ≤ n ≤ 22; <image>) — the number of guests at the party (including Arseny) and the number of pairs of people which are friends.

Each of the next m lines contains two integers u and v (1 ≤ u, v ≤ n; u ≠ v), which means that people with numbers u and v are friends initially. It's guaranteed that each pair of friends is described not more than once and the graph of friendship is connected.

Output

In the first line print the minimum number of steps required to make all pairs of guests friends.

In the second line print the ids of guests, who are selected at each step.

If there are multiple solutions, you can output any of them.

Examples

Input

5 6
1 2
1 3
2 3
2 5
3 4
4 5


Output

2
2 3 

Input

4 4
1 2
1 3
1 4
3 4


Output

1
1 

Note

In the first test case there is no guest who is friend of all other guests, so at least two steps are required to perform the task. After second guest pairwise introduces all his friends, only pairs of guests (4, 1) and (4, 2) are not friends. Guest 3 or 5 can introduce them.

In the second test case guest number 1 is a friend of all guests, so he can pairwise introduce all guests in one step.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
std::istream& operator>>(std::istream& i, pair<T, U>& p) {
  i >> p.first >> p.second;
  return i;
}
template <typename T>
std::istream& operator>>(std::istream& i, vector<T>& t) {
  for (auto& v : t) {
    i >> v;
  }
  return i;
}
template <typename T>
std::ostream& operator<<(std::ostream& o, const vector<T>& t) {
  for (size_t i = 0; i < t.size(); ++i) {
    o << t[i] << " \n"[i == t.size() - 1];
  }
  return o;
}
template <typename T>
using minheap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using maxheap = priority_queue<T, vector<T>, less<T>>;
auto fraclt = [](const std::pair<int, int>& a, const std::pair<int, int>& b) {
  return (long long)a.first * b.second < (long long)b.first * a.second;
};
struct cmpfrac {
  bool operator()(const std::pair<int, int>& a,
                  const std::pair<int, int>& b) const {
    return (long long)a.first * b.second < (long long)b.first * a.second;
  }
};
template <typename T>
bool in(T a, T b, T c) {
  return a <= b && b < c;
}
unsigned int logceil(long long first) {
  unsigned int b = 0;
  while (first) {
    first >>= 1;
    ++b;
  }
  return b;
}
namespace std {
template <typename T, typename U>
struct hash<pair<T, U>> {
  hash<T> t;
  hash<U> u;
  size_t operator()(const pair<T, U>& p) const {
    return t(p.first) ^ (u(p.second) << 7);
  }
};
}  // namespace std
template <typename T, typename F>
T bsh(T l, T h, const F& f) {
  T r = -1, m;
  while (l <= h) {
    m = (l + h) / 2;
    if (f(m)) {
      l = m + 1;
      r = m;
    } else {
      h = m - 1;
    }
  }
  return r;
}
template <typename F>
double bsh(double l, double h, const F& f, double p = 1e-9) {
  unsigned int r = 3 + (unsigned int)log2((h - l) / p);
  while (r--) {
    double m = (l + h) / 2;
    if (f(m)) {
      l = m;
    } else {
      h = m;
    }
  }
  return (l + h) / 2;
}
template <typename T, typename F>
T bsl(T l, T h, const F& f) {
  T r = -1, m;
  while (l <= h) {
    m = (l + h) / 2;
    if (f(m)) {
      h = m - 1;
      r = m;
    } else {
      l = m + 1;
    }
  }
  return r;
}
template <typename F>
double bsl(double l, double h, const F& f, double p = 1e-9) {
  unsigned int r = 3 + (unsigned int)log2((h - l) / p);
  while (r--) {
    double m = (l + h) / 2;
    if (f(m)) {
      h = m;
    } else {
      l = m;
    }
  }
  return (l + h) / 2;
}
template <typename T>
T gcd(T a, T b) {
  if (a < b) swap(a, b);
  return b ? gcd(b, a % b) : a;
}
template <typename T>
class vector2 : public vector<vector<T>> {
 public:
  vector2() {}
  vector2(size_t a, size_t b, T t = T())
      : vector<vector<T>>(a, vector<T>(b, t)) {}
};
template <typename T>
class vector3 : public vector<vector2<T>> {
 public:
  vector3() {}
  vector3(size_t a, size_t b, size_t c, T t = T())
      : vector<vector2<T>>(a, vector2<T>(b, c, t)) {}
};
template <typename T>
class vector4 : public vector<vector3<T>> {
 public:
  vector4() {}
  vector4(size_t a, size_t b, size_t c, size_t d, T t = T())
      : vector<vector3<T>>(a, vector3<T>(b, c, d, t)) {}
};
template <typename T>
class vector5 : public vector<vector4<T>> {
 public:
  vector5() {}
  vector5(size_t a, size_t b, size_t c, size_t d, size_t e, T t = T())
      : vector<vector4<T>>(a, vector4<T>(b, c, d, e, t)) {}
};
template <typename T>
struct bounded_priority_queue {
  inline bounded_priority_queue(unsigned int X) : A(X), B(0) {}
  inline void push(unsigned int L, T V) {
    B = max(B, L);
    A[L].push(V);
  }
  inline const T& top() const { return A[B].front(); }
  inline void pop() {
    A[B].pop();
    while (B > 0 && A[B].empty()) --B;
  }
  inline bool empty() const { return A[B].empty(); }
  inline void clear() {
    B = 0;
    for (auto& a : A) a = queue<T>();
  }

 private:
  vector<queue<T>> A;
  unsigned int B;
};
class TaskC {
 public:
  int N, M;
  vector<int> E;
  void solve(istream& cin, ostream& cout) {
    cin >> N >> M;
    if (M == N * (N - 1) / 2) {
      cout << "0\n";
      return;
    }
    E = vector<int>(N, 0);
    for (int i = 0; i < M; ++i) {
      int a, b;
      cin >> a >> b;
      --a;
      --b;
      E[a] |= (1 << b);
      E[b] |= (1 << a);
    }
    vector<int> D(1 << N, 0);
    for (int i = 0; i < N; ++i) {
      E[i] |= (1 << i);
      D[1 << i] = E[i];
    }
    int all = (1 << N) - 1;
    int b = all, c = N;
    for (int i = 1; i < (1 << N); ++i) {
      if (D[i] == all) {
        int e = __builtin_popcount(i);
        if (e < c) {
          b = i;
          c = e;
        }
      } else if (D[i] != 0) {
        for (int j = 0; j < N; ++j) {
          if ((D[i] & (1 << j)) == (1 << j) && (i & (1 << j)) == 0) {
            D[i | (1 << j)] = D[i] | E[j];
          }
        }
      }
    }
    cout << c << endl;
    for (int i = 0; i < N; ++i) {
      if ((b & (1 << i)) == (1 << i)) {
        cout << i + 1 << ' ';
      }
    }
    cout << endl;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  TaskC solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}