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 N = 100000;
unordered_set<int> adj[N];
bool visited[N];
int degree[N];
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  fill(adj, adj + n, unordered_set<int>());
  fill(visited, visited + n, false);
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    adj[u].insert(v);
    adj[v].insert(u);
  }
  set<array<int, 2>> nodes;
  for (int i = 0; i < n; ++i) {
    degree[i] = adj[i].size();
    nodes.insert({degree[i], i});
  }
  vector<vector<int>> todo;
  while (!nodes.empty()) {
    auto [d, u] = *nodes.begin();
    nodes.erase(nodes.begin());
    if (d == k - 1) {
      vector<int> v = {u};
      for (auto c : adj[u]) {
        if (!visited[c]) {
          v.push_back(c);
        }
      }
      todo.push_back(v);
    } else if (d >= k) {
      cout << 1 << " " << count(visited, visited + n, false) << "\n";
      for (int i = 0; i < n; ++i) {
        if (!visited[i]) {
          cout << i + 1 << " ";
        }
      }
      cout << "\n";
      return;
    }
    visited[u] = true;
    for (auto c : adj[u]) {
      if (!visited[c]) {
        nodes.erase({degree[c], c});
        nodes.insert({--degree[c], c});
      }
    }
  }
  for (auto &v : todo) {
    bool ok = true;
    for (int i = 0; i < (int)v.size() && ok; ++i) {
      for (int j = i + 1; j < (int)v.size() && ok; ++j) {
        ok &= adj[v[i]].count(v[j]) > 0;
      }
    }
    if (ok) {
      cout << 2 << "\n";
      for (auto i : v) {
        cout << i + 1 << " ";
      }
      cout << "\n";
      return;
    }
  }
  cout << -1 << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}