Problem: Given a connected undirected graph with n vertices and an integer k, you have to either:

  * either find an independent set that has exactly ⌈k/2⌉ vertices.
  * or find a simple cycle of length at most k. 



An independent set is a set of vertices such that no two of them are connected by an edge. A simple cycle is a cycle that doesn't contain any vertex twice. 

I have a proof that for any input you can always solve at least one of these problems, but it's left as an exercise for the reader.

Input

The first line contains three integers n, m, and k (3 ≤ k ≤ n ≤ 10^5, n-1 ≤ m ≤ 2 ⋅ 10^5) — the number of vertices and edges in the graph, and the parameter k from the statement.

Each of the next m lines contains two integers u and v (1 ≤ u,v ≤ n) that mean there's an edge between vertices u and v. It's guaranteed that the graph is connected and doesn't contain any self-loops or multiple edges.

Output

If you choose to solve the first problem, then on the first line print 1, followed by a line containing ⌈k/2⌉ distinct integers not exceeding n, the vertices in the desired independent set.

If you, however, choose to solve the second problem, then on the first line print 2, followed by a line containing one integer, c, representing the length of the found cycle, followed by a line containing c distinct integers not exceeding n, the vertices in the desired cycle, in the order they appear in the cycle.

Examples

Input


4 4 3
1 2
2 3
3 4
4 1


Output


1
1 3 

Input


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


Output


2
3
2 3 4 

Input


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


Output


2
3
1 2 3 

Input


5 4 5
1 2
1 3
2 4
2 5


Output


1
1 4 5 

Note

In the first sample:

<image>

Notice that printing the independent set \{2,4\} is also OK, but printing the cycle 1-2-3-4 isn't, because its length must be at most 3.

In the second sample:

<image>

Notice that printing the independent set \{1,3\} or printing the cycle 2-1-4 is also OK.

In the third sample:

<image>

In the fourth sample:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void _R(T &x) {
  cin >> x;
}
void _R(int &x) { scanf("%d", &x); }
void _R(long long &x) { cin >> x; }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template <typename T, typename... U>
void R(T &head, U &...tail) {
  _R(head);
  R(tail...);
}
template <typename T>
void _W(const T &x) {
  cout << x;
}
void _W(const int &x) { printf("%d", x); }
template <typename T>
void _W(const vector<T> &x) {
  for (auto i = x.cbegin(); i != x.cend(); i++) {
    if (i != x.cbegin()) putchar(' ');
    _W(*i);
  }
}
void W() {}
template <typename T, typename... U>
void W(const T &head, const U &...tail) {
  _W(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  W(tail...);
}
template <class T>
inline void maz(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
inline void miz(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
const int N = 1e5 + 10;
const int INF = 1e9 + 7;
const int MOD = 1e9 + 7;
int n, m;
int sqn;
vector<int> G[N];
bool mis() {
  static int deg[N];
  static int sta[N];
  int cnt = 0;
  for (int i = 0; i < n; i++) deg[i] = G[i].size();
  priority_queue<std::pair<int, int> > Q;
  for (int i = 0; i < n; i++) Q.push(std::make_pair(-deg[i], i));
  while (!Q.empty()) {
    const auto t = Q.top();
    Q.pop();
    const int x = t.second, d = -t.first;
    if (deg[x] != d) continue;
    if (sta[x] != 0) continue;
    sta[x] = 1;
    cnt++;
    if (cnt == sqn) break;
    for (auto y : G[x]) {
      if (sta[y] != -1) {
        assert(sta[y] == 0);
        sta[y] = -1;
        for (auto z : G[y])
          if (sta[z] == 0) {
            --deg[z];
            Q.push(std::make_pair(-deg[z], z));
          }
      }
    }
  }
  if (cnt != sqn) return false;
  puts("1");
  for (int i = 0; i < n; i++)
    if (sta[i] == 1) printf("%d ", i + 1);
  puts("");
  return true;
}
int vis[N], vtop;
int par[N], dep[N];
void print(int x, int y) {
  puts("2");
  printf("%d\n", dep[x] - dep[y] + 1);
  while (x != y) {
    printf("%d ", x + 1);
    x = par[x];
  }
  printf("%d\n", y + 1);
}
void dfs(int x, int d) {
  dep[x] = d;
  vis[x] = vtop;
  for (auto y : G[x])
    if (vis[y] == vtop) {
      if (dep[x] - dep[y] + 1 >= sqn) {
        print(x, y);
        exit(0);
      }
    } else {
      par[y] = x;
      dfs(y, d + 1);
    }
}
bool find_cycle(int rt) {
  vtop++;
  dfs(rt, 0);
  return false;
}
int main() {
  R(n, m);
  for (int i = 0; i < m; i++) {
    int x, y;
    R(x, y);
    x--;
    y--;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  sqn = sqrt(n);
  while (sqn * sqn < n) sqn++;
  if (mis()) return 0;
  int T = 1;
  vector<int> V;
  for (int i = 0; i < n; i++) V.push_back(i);
  srand(time(NULL));
  std::random_device rd;
  std::mt19937 g(rd());
  shuffle((V).begin(), (V).end(), g);
  while (T--)
    if (find_cycle(V[T])) break;
  return 0;
}