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 auto maxn = int(1e5 + 5);
inline auto FastIO() -> void {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
auto g = vector<vector<int>>(maxn);
auto degree = vector<int>(maxn);
auto visited = vector<bool>(maxn);
auto drop = vector<bool>(maxn);
auto clique = vector<int>(maxn);
auto subset = vector<int>(maxn);
auto n = 0, m = 0, k = 0;
auto ans = -1;
auto preprocess() -> void {
  cin >> n >> m >> k;
  for (auto i = 1; i <= n; ++i) {
    g[i].clear();
    degree[i] = 0;
    drop[i] = visited[i] = false;
  }
  auto x = 0, y = 0;
  for (auto i = 0; i < m; ++i) {
    cin >> x >> y;
    degree[x]++;
    degree[y]++;
    g[x].emplace_back(y);
    g[y].emplace_back(x);
  }
  for (auto i = 1; i <= n; ++i) sort(g[i].begin(), g[i].end());
}
auto solve() -> void {
  if (1ll * k * (k - 1) > 2 * m) {
    ans = -1;
    return;
  }
  auto q = queue<int>();
  for (auto i = 1; i <= n; ++i) {
    if (degree[i] < k) {
      visited[i] = true;
      q.push(i);
    }
  }
  while (!q.empty()) {
    auto cur = q.front();
    q.pop();
    if (degree[cur] == k - 1) {
      clique.clear();
      clique.emplace_back(cur);
      for (auto v : g[cur]) {
        if (!drop[v]) clique.emplace_back(v);
      }
      bool hasClique = true;
      for (auto u : clique) {
        for (auto v : clique) {
          if (u == v) break;
          if (!binary_search(g[u].begin(), g[u].end(), v)) hasClique = false;
        }
      }
      if (hasClique) {
        ans = 2;
        return;
      }
    }
    drop[cur] = true;
    for (auto v : g[cur]) {
      --degree[v];
      if (degree[v] < k && !visited[v]) {
        visited[v] = true;
        q.push(v);
      }
    }
  }
  ans = -1;
  subset.clear();
  for (auto i = 1; i <= n; ++i) {
    if (!visited[i]) {
      subset.push_back(i);
      ans = 1;
    }
  }
}
auto print = [](int x) { cout << " " << x; };
auto printAns() -> void {
  if (ans == -1) {
    cout << -1 << endl;
  } else if (ans == 1) {
    cout << ans << " " << subset.size() << endl;
    cout << subset.front();
    for_each(next(subset.begin()), subset.end(), print);
    cout << endl;
  } else if (ans == 2) {
    cout << ans << endl;
    cout << clique.front();
    for_each(next(clique.begin()), clique.end(), print);
    cout << endl;
  }
}
auto main() -> int {
  FastIO();
  auto t = 0;
  cin >> t;
  while (t-- > 0) {
    preprocess();
    solve();
    printAns();
  }
  return 0;
}