Problem: You are given an undirected graph with n vertices and m edges. Also, you are given an integer k.

Find either a clique of size k or a non-empty subset of vertices such that each vertex of this subset has at least k neighbors in the subset. If there are no such cliques and subsets report about it.

A subset of vertices is called a clique of size k if its size is k and there exists an edge between every two vertices from the subset. A vertex is called a neighbor of the other vertex if there exists an edge between them.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The next lines contain descriptions of test cases.

The first line of the description of each test case contains three integers n, m, k (1 ≤ n, m, k ≤ 10^5, k ≤ n).

Each of the next m lines contains two integers u, v (1 ≤ u, v ≤ n, u ≠ v), denoting an edge between vertices u and v.

It is guaranteed that there are no self-loops or multiple edges. It is guaranteed that the sum of n for all test cases and the sum of m for all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case: 

If you found a subset of vertices such that each vertex of this subset has at least k neighbors in the subset in the first line output 1 and the size of the subset. On the second line output the vertices of the subset in any order.

If you found a clique of size k then in the first line output 2 and in the second line output the vertices of the clique in any order.

If there are no required subsets and cliques print -1.

If there exists multiple possible answers you can print any of them.

Example

Input


3
5 9 4
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
10 15 3
1 2
2 3
3 4
4 5
5 1
1 7
2 8
3 9
4 10
5 6
7 10
10 8
8 6
6 9
9 7
4 5 4
1 2
2 3
3 4
4 1
1 3


Output


2
4 1 2 3 
1 10
1 2 3 4 5 6 7 8 9 10 
-1

Note

In the first test case: the subset \{1, 2, 3, 4\} is a clique of size 4.

In the second test case: degree of each vertex in the original graph is at least 3. So the set of all vertices is a correct answer.

In the third test case: there are no cliques of size 4 or required subsets, so the answer is -1.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& it : v) is >> it;
  return is;
}
template <class L, class R>
ostream& operator<<(ostream& os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <class T>
ostream& operator<<(ostream& os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << "]";
}
template <class T>
ostream& operator<<(ostream& os, set<T> second) {
  os << "{ ";
  for (auto s : second) os << s << " ";
  return os << "}";
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " |";
  __f(comma + 1, args...);
}
const int N = 1e5;
unordered_set<int> adj[N];
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 0; i < n; ++i) {
    adj[i].clear();
  }
  vector<int> degree(n);
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    adj[u].insert(v);
    adj[v].insert(u);
    degree[u]++;
    degree[v]++;
  }
  set<pair<int, int> > degreePair;
  for (int i = 0; i < n; ++i) {
    degreePair.insert({degree[i], i});
  }
  while (degreePair.size()) {
    auto small = *degreePair.begin();
    int node = small.second, deg = small.first;
    if (deg >= k) {
      cout << 1 << " " << degreePair.size() << '\n';
      for (pair<int, int> p : degreePair) {
        cout << p.second + 1 << " ";
      }
      cout << '\n';
      return;
    }
    if (deg == k - 1) {
      vector<int> clique{node};
      for (int child : adj[node]) {
        clique.push_back(child);
      }
      bool isClique = true;
      for (int i = 0; i < k; ++i) {
        int u = clique[i];
        for (int j = 0; j < i; ++j) {
          int v = clique[j];
          if (adj[u].count(v) == 0) {
            isClique = false;
            break;
          }
        }
        if (isClique == false) break;
      }
      if (isClique) {
        cout << 2 << '\n';
        for (int i : clique) cout << i + 1 << " ";
        cout << '\n';
        return;
      }
    }
    degreePair.erase(small);
    for (int par : adj[node]) {
      degreePair.erase({degree[par], par});
      degree[par]--;
      adj[par].erase(node);
      degreePair.insert({degree[par], par});
    }
    adj[node].clear();
    degree[node] = 0;
  }
  cout << "-1\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}