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;
int T, n, m, k;
unordered_set<int> g[100003];
int v[100003];
vector<int> ans;
int sol() {
  ans.clear();
  priority_queue<pair<int, int> > q;
  for (int i = 1; i < n + 1; ++i) {
    v[i] = 0;
    q.push(make_pair(-g[i].size(), i));
  }
  while (!q.empty()) {
    auto t = q.top();
    q.pop();
    int u = t.second;
    if (v[u]) continue;
    if (g[u].size() >= k) {
      for (int i = 1; i <= n; ++i) {
        if (v[i]) continue;
        ans.push_back(i);
      }
      return 1;
    } else if (g[u].size() == k - 1 && ((long long)k * (k - 1ll)) / 2 <= m) {
      int ok = 1;
      for (auto it = g[u].begin(); it != g[u].end() && ok; ++it) {
        auto itt = it;
        for (itt++; itt != g[u].end(); ++itt) {
          if (g[*it].find(*itt) == g[*it].end()) {
            ok = 0;
            break;
          }
        }
      }
      if (ok) {
        ans.push_back(u);
        for (auto it : g[u]) {
          ans.push_back(it);
        }
        return 2;
      }
    }
    v[u] = 1;
    for (auto it : g[u]) {
      g[it].erase(u);
      q.push(make_pair(-g[it].size(), it));
    }
  }
  return 0;
}
int main() {
  ios::sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> m >> k;
    for (int i = 0; i < n + 1; ++i) {
      g[i].clear();
    }
    for (int i = 0; i < m; ++i) {
      int u, v;
      cin >> u >> v;
      g[u].insert(v);
      g[v].insert(u);
    }
    int ret = sol();
    if (ret == 1) {
      cout << 1 << " " << ans.size() << endl;
    } else if (ret == 2) {
      cout << 2 << endl;
    } else {
      cout << -1 << endl;
      continue;
    }
    for (int t : ans) {
      cout << t << " ";
    }
    cout << endl;
  }
  return 0;
}