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;
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<vector<int> > adj(n);
  vector<int> deg(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());
  }
  set<pair<int, int> > st;
  for (int i = 0; i < n; i++) {
    st.insert(make_pair(adj[i].size(), i));
  }
  vector<bool> delet(n);
  vector<vector<int> > candidates;
  while (!st.empty()) {
    int dg = st.begin()->first;
    int u = st.begin()->second;
    if (dg >= k) break;
    if (dg == k - 1) {
      vector<int> v_;
      v_.push_back(u);
      for (int i = 0; i < adj[u].size(); i++) {
        int v = adj[u][i];
        if (!delet[v]) v_.push_back(v);
      }
      if (v_.size() == k) candidates.push_back(v_);
    }
    st.erase(st.begin());
    delet[u] = true;
    for (int i = 0; i < adj[u].size(); i++) {
      int v = adj[u][i];
      if (!delet[v]) {
        st.erase(st.find(make_pair(deg[v], v)));
        deg[v]--;
        st.insert(make_pair(deg[v], v));
      }
    }
  }
  if (st.empty()) {
    for (int i = 0; i < candidates.size(); i++) {
      vector<int> v_ = candidates[i];
      int sz = v_.size();
      bool ok = true;
      for (int j = 0; j < v_.size() - 1; j++) {
        for (int k = j + 1; k < v_.size(); k++) {
          int u = v_[j];
          int v = v_[k];
          if (!binary_search(adj[u].begin(), adj[u].end(), v)) {
            ok = false;
            break;
          }
        }
        if (!ok) break;
      }
      if (ok) {
        cout << "2\n";
        for (int j = 0; j < v_.size(); j++) {
          cout << v_[j] + 1 << ' ';
        }
        cout << '\n';
        return;
      }
    }
    cout << "-1\n";
    return;
  } else {
    cout << "1 " << st.size() << '\n';
    for (set<pair<int, int> >::iterator it = st.begin(); it != st.end(); it++) {
      int v = it->second;
      cout << v + 1 << ' ';
    }
    cout << '\n';
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}