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 <class T, class U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "{";
  for (int i = 0; i < (int)v.size(); i++) {
    if (i) os << ", ";
    os << v[i];
  }
  os << "}";
  return os;
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  srand(time(NULL));
  while (t--) {
    int n, m, k;
    cin >> n >> m >> k;
    set<pair<int, int>> nodes;
    vector<vector<int>> adj(n);
    vector<int> deg(n);
    vector<int> removed(n);
    for (int i = 0; i < m; i++) {
      int u, v;
      cin >> u >> v;
      u--;
      v--;
      adj[u].push_back(v);
      adj[v].push_back(u);
      deg[u]++;
      deg[v]++;
    }
    for (int i = 0; i < n; i++) sort(((adj[i]).begin()), ((adj[i]).end()));
    for (int i = 0; i < n; i++) {
      nodes.insert({deg[i], i});
    }
    bool done = false;
    while (!nodes.empty()) {
      auto it = nodes.begin();
      int u = it->second;
      int d = it->first;
      nodes.erase(it);
      removed[u] = 1;
      vector<int> vec;
      for (int v : adj[u])
        if (!removed[v]) vec.push_back(v);
      random_shuffle(((vec).begin()), ((vec).end()));
      if (d + 1 < k) {
        for (int v : vec) {
          nodes.erase({deg[v], v});
          deg[v]--;
          nodes.insert({deg[v], v});
        }
      } else if (d + 1 == k) {
        bool clique = true;
        for (int i = 0; i < ((int)(vec).size()) && clique; i++) {
          for (int j = 0; j < i; j++) {
            int p = vec[i], q = vec[j];
            auto it2 = upper_bound(((adj[p]).begin()), ((adj[p]).end()), q - 1);
            if (it2 == adj[p].end() || *it2 != q) {
              clique = false;
              break;
            }
          }
        }
        if (clique) {
          cout << 2 << "\n";
          cout << u + 1 << " ";
          for (int v : vec) cout << v + 1 << " ";
          cout << "\n";
          done = true;
          break;
        } else {
          for (int v : vec) {
            nodes.erase({deg[v], v});
            deg[v]--;
            nodes.insert({deg[v], v});
          }
        }
      } else {
        cout << "1 "
             << n + 1 - accumulate(((removed).begin()), ((removed).end()), 0)
             << "\n";
        cout << u + 1 << " ";
        for (int i = 0; i < n; i++)
          if (!removed[i]) cout << i + 1 << " ";
        cout << "\n";
        done = true;
        break;
      }
    }
    if (!done) cout << "-1\n";
  }
}