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>> g(n);
  vector<int> sz(n), used(n);
  unordered_set<long long> con;
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    g[a].push_back(b);
    g[b].push_back(a);
    sz[a]++, sz[b]++;
    con.insert(a * n + b);
    con.insert(b * n + a);
  }
  set<pair<int, int>> st;
  for (int i = 0; i < n; ++i) st.insert({(int)g[i].size(), i});
  vector<vector<int>> lol;
  while (!st.empty() && st.begin()->first < k) {
    int v = st.begin()->second;
    used[v] = 1;
    st.erase(st.begin());
    if (sz[v] == k - 1) {
      vector<int> kek = {v};
      for (int j : g[v])
        if (!used[j]) kek.push_back(j);
      lol.push_back(kek);
    }
    for (int i : g[v]) {
      if (!used[i]) {
        st.erase({sz[i], i});
        sz[i]--;
        st.insert({sz[i], i});
      }
    }
  }
  if (!st.empty()) {
    cout << "1 " << st.size() << '\n';
    for (auto &i : st) cout << i.second + 1 << ' ';
    cout << '\n';
    return;
  }
  for (auto &comp : lol) {
    bool ok = true;
    for (int i = 0; i < comp.size() && ok; ++i)
      for (int j = i + 1; j < comp.size() && ok; ++j)
        if (con.find(comp[i] * n + comp[j]) == con.end()) ok = false;
    if (ok) {
      cout << "2\n";
      for (int j : comp) cout << j + 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();
  }
}