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 tam = 100010;
const int MOD = 1000000007;
const int MOD1 = 998244353;
const double EPS = 1e-9;
const double PI = acos(-1);
void solve() {
  int n, m, k, a, b;
  cin >> n >> m >> k;
  vector<unordered_set<int>> ma(n + 1);
  vector<vector<int>> g(n + 1);
  vector<int> deg(n + 1);
  vector<bool> active(n + 1, true);
  while (m--) {
    cin >> a >> b;
    a--;
    b--;
    ma[a].insert(b);
    ma[b].insert(a);
    g[a].push_back(b);
    g[b].push_back(a);
    deg[a]++;
    deg[b]++;
  }
  set<pair<int, int>> st;
  for (int i = 0; i < n; i++) st.insert({deg[i], i});
  int key;
  while (st.size()) {
    key = st.begin()->second;
    st.erase(*st.begin());
    active[key] = 0;
    vector<int> vec;
    if (deg[key] >= k) {
      for (int i = 0; i < n; i++)
        if (active[i] || key == i) vec.push_back(i);
      cout << "1 " << (int)vec.size() << "\n";
      for (int i = 0; i < vec.size(); i++) {
        if (i > 0) cout << ' ';
        cout << vec[i] + 1;
      }
      cout << '\n';
      return;
    } else if (deg[key] == k - 1) {
      for (int x : g[key]) {
        if (active[x]) vec.push_back(x);
      }
      bool bo = true;
      for (int i = 0; i < k - 1; i++)
        for (int j = i + 1; j < k - 1; j++) {
          if (!ma[vec[i]].count(vec[j])) {
            bo = false;
            i = MOD;
            j = MOD;
          }
        }
      if (bo) {
        cout << "2\n";
        cout << key + 1;
        for (int i = 0; i < k - 1; i++) {
          cout << ' ' << vec[i] + 1;
        }
        cout << '\n';
        return;
      }
    }
    for (int x : g[key]) {
      if (active[x]) {
        st.erase({deg[x], x});
        deg[x]--;
        st.insert({deg[x], x});
      }
    }
  }
  cout << "-1\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int cas;
  cin >> cas;
  while (cas--) {
    solve();
  }
  return 0;
}