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;
int n, m, k;
unordered_set<int> G[100013];
int deg[100013];
vector<int> r2;
bool ck(int u) {
  for (auto i = G[u].begin(); i != G[u].end(); i++) {
    for (auto j = i; j != G[u].end(); j++) {
      if (i == j) continue;
      if (G[*i].find(*j) == G[*i].end()) return 0;
    }
  }
  r2.clear();
  r2.push_back(u);
  for (auto i : G[u]) {
    r2.push_back(i);
  }
  return 1;
}
priority_queue<pair<int, int>> q;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T, de, u, v;
  cin >> T;
  while (T--) {
    cin >> n >> m >> k;
    while (!q.empty()) q.pop();
    for (int i = 1; i <= n; i++) {
      G[i].clear();
    }
    for (int i = 0; i < m; i++) {
      cin >> u >> v;
      G[u].insert(v);
      G[v].insert(u);
    }
    if ((k - 1ll) * k > 2ll * m) {
      cout << "-1\n";
      continue;
    }
    for (int i = 1; i <= n; i++) {
      deg[i] = (int)G[i].size();
      q.push(make_pair(-deg[i], i));
    }
    int res = -1;
    while (!q.empty()) {
      de = -q.top().first, u = q.top().second;
      if (!deg[u]) {
        q.pop();
        continue;
      }
      if (de >= k) {
        res = 1;
        break;
      }
      if (de == k - 1 && ck(u)) {
        res = 2;
        break;
      }
      q.pop();
      for (auto v : G[u]) {
        G[v].erase(u);
        deg[v]--;
        q.push(make_pair(-deg[v], v));
      }
      G[u].clear(), deg[u] = 0;
    }
    if (res == 1) {
      r2.clear();
      for (int i = 1; i <= n; i++) {
        if (deg[i]) r2.push_back(i);
      }
      int si = (int)r2.size();
      cout << "1 " << si << "\n";
      for (int i = 0; i < si; i++) cout << r2[i] << " ";
      cout << "\n";
    } else if (res == 2) {
      cout << "2\n";
      for (int i = 0; i < k; i++) cout << r2[i] << " ";
      cout << "\n";
    } else
      cout << "-1\n";
  }
  return 0;
}