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;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const long long inf = 1e9 + 7;
const int max_n = 1e5 + 3;
set<int> gr[max_n];
vector<int> gr1[max_n];
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) {
    gr[i].clear();
    gr1[i].clear();
  }
  vector<int> deg(n);
  auto sorti = [&deg](int v1, int v2) {
    if (deg[v1] == deg[v2]) return v1 < v2;
    return deg[v1] < deg[v2];
  };
  set<int, decltype(sorti)> a(sorti);
  for (int i = 0; i < m; i++) {
    int v1, v2;
    cin >> v1 >> v2;
    v1--;
    v2--;
    gr[v1].emplace(v2);
    gr[v2].emplace(v1);
    gr1[v1].emplace_back(v2);
    gr1[v2].emplace_back(v1);
    deg[v1]++;
    deg[v2]++;
  }
  for (int i = 0; i < n; i++) sort(gr1[i].begin(), gr1[i].end());
  for (int i = 0; i < n; i++) a.emplace(i);
  vector<bool> fuck(n);
  while ((int)(a.size())) {
    int v = (*a.begin());
    a.erase(a.begin());
    if (deg[v] >= k) {
      a.emplace(v);
      cout << 1 << " " << (int)(a.size()) << endl;
      for (int i : a) {
        cout << i + 1 << " ";
      }
      cout << endl;
      return;
    }
    if (!fuck[v] && deg[v] == k - 1) {
      bool bad = false;
      for (int v1 : gr[v]) {
        for (int v2 : gr[v]) {
          if (v1 == v2) break;
          int id =
              lower_bound(gr1[v1].begin(), gr1[v1].end(), v2) - gr1[v1].begin();
          if (id == (int)(gr1[v1].size()) || gr1[v1][id] != v2) bad = true;
        }
        if (bad) break;
      }
      if (bad) {
        for (int i : gr[v]) {
          if (deg[i] == k - 1) fuck[i] = true;
        }
      } else {
        cout << 2 << endl;
        for (int i : gr[v]) cout << i + 1 << " ";
        cout << v + 1 << endl;
        return;
      }
    }
    for (int i : gr[v]) {
      a.erase(i);
      gr[i].erase(v);
      deg[i]--;
      a.emplace(i);
    }
  }
  cout << -1 << endl;
}
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) solve();
}