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;
const int MxN = (int)1e5 + 9;
unordered_set<int> adj[MxN + 9];
queue<int> q;
int n, m, k;
void enqueue(const int);
void rem(int u) {
  assert(adj[u].size() <= k - 1);
  vector<int> adj_u(adj[u].begin(), adj[u].end());
  adj[u].clear();
  for (const int v : adj_u) {
    auto it = adj[v].find(u);
    if (it == adj[v].end()) continue;
    adj[v].erase(it);
    enqueue(v);
  }
}
void enqueue(const int u) {
  if (adj[u].size() < k - 1)
    rem(u);
  else if (adj[u].size() == k - 1)
    q.push(u);
}
void f() {
  cin >> n >> m >> k;
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].insert(v);
    adj[v].insert(u);
  }
  if (k * (k - 1LL) > 2 * m) {
    cout << "-1\n";
    return;
  }
  for (int i = 1; i <= n; ++i) enqueue(i);
  while (!q.empty()) {
    const int u = q.front();
    q.pop();
    assert(adj[u].size() <= k - 1);
    if (adj[u].size() < k - 1) {
      rem(u);
      continue;
    }
    bool ok = true;
    for (auto it1 = adj[u].begin(); it1 != adj[u].end() && ok; ++it1) {
      for (auto it2 = next(it1); it2 != adj[u].end() && ok; ++it2) {
        if (!adj[*it1].count(*it2)) {
          ok = false;
          break;
        }
      }
    }
    if (ok) {
      cout << "2\n" << u;
      for (int x : adj[u]) cout << ' ' << x;
      cout << '\n';
      return;
    }
    rem(u);
  }
  vector<int> v;
  for (int i = 1; i <= n; ++i)
    if (!adj[i].empty()) {
      v.push_back(i);
      assert(adj[i].size() >= k);
    }
  if (v.empty()) {
    cout << "-1\n";
    return;
  }
  cout << "1 " << v.size() << '\n';
  for (int x : v) cout << ' ' << x;
  cout << '\n';
  return;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    f();
    for (int i = 0; i <= n; ++i) {
      adj[i].clear();
    }
    q = queue<int>{};
  }
  return 0;
}