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 T, n, m, k;
unordered_set<int> g[100003];
vector<int> ans;
int sol() {
  bool kans = 0;
  ans.clear();
  priority_queue<pair<int, int> > q;
  for (int i = 1; i <= n; ++i) {
    q.push(make_pair(-g[i].size(), i));
  }
  while (!q.empty()) {
    int t = q.top().second;
    q.pop();
    if (g[t].empty()) continue;
    if (g[t].size() >= k) {
      kans = 1;
      break;
    }
    if (g[t].size() == k - 1 && k * (k - 1ll) / 2 <= m) {
      bool f = 1;
      for (auto p = g[t].begin(); f && p != g[t].end(); ++p)
        for (auto q = p; f && ++q != g[t].end();)
          f &= g[*p].find(*q) != g[*p].end();
      if (f) {
        ans.push_back(t);
        ans.insert(ans.end(), g[t].begin(), g[t].end());
        break;
      }
    }
    for (auto p : g[t]) {
      g[p].erase(t);
      q.push(make_pair(-g[p].size(), p));
    }
    g[t].clear();
  }
  if (ans.size()) {
    return 2;
  } else if (kans) {
    for (int i = 1; i <= n; ++i)
      if (g[i].size()) ans.push_back(i);
    return 1;
  }
  return -1;
}
int main() {
  ios::sync_with_stdio(0);
  for (cin >> T; T; --T) {
    cin >> n >> m >> k;
    for (int i = 1; i <= n; ++i) g[i].clear();
    for (int i = 1; i <= m; ++i) {
      int p, q;
      cin >> p >> q;
      g[p].insert(q);
      g[q].insert(p);
    }
    int ret = sol();
    if (ret == 2) {
      cout << 2 << '\n';
      for (auto t : ans) cout << t << ' ';
      cout << endl;
    } else if (ret == 1) {
      cout << 1 << ' ' << ans.size() << '\n';
      for (auto t : ans) cout << t << ' ';
      cout << endl;
    } else
      cout << -1 << endl;
  }
  return 0;
}