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 rng(chrono::steady_clock::now().time_since_epoch().count());
int randi = uniform_int_distribution<int>(0, 10)(rng);
const int N = 1e5 + 5;
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  unordered_set<int> g[n];
  unordered_set<int> g2[n];
  for (int i = 0; i < m; i++) {
    int v, u;
    cin >> v >> u;
    v--, u--;
    g[v].insert(u);
    g[u].insert(v);
    g2[v].insert(u);
    g2[u].insert(v);
  }
  set<pair<int, int>> s;
  for (int i = 0; i < n; i++) {
    s.insert({g2[i].size(), i});
  }
  vector<vector<int>> cands;
  while (!s.empty()) {
    auto fi = s.begin();
    int v = fi->second;
    s.erase(fi);
    if (g2[v].size() < k - 1) {
      for (const int &u : g2[v]) {
        s.erase({g2[u].size(), u});
        g2[u].erase(v);
        s.insert({g2[u].size(), u});
      }
    } else if (g2[v].size() == k - 1) {
      vector<int> c;
      c.reserve(k);
      c.push_back(v);
      for (const int &u : g2[v]) {
        c.push_back(u);
        s.erase({g2[u].size(), u});
        g2[u].erase(v);
        s.insert({g2[u].size(), u});
      }
      cands.push_back(c);
    } else {
      cout << 1 << " " << (1 + s.size()) << "\n";
      cout << (v + 1) << " ";
      for (const auto &p : s) {
        cout << (p.second + 1) << " ";
      }
      cout << "\n";
      return;
    }
  }
  for (const vector<int> &c : cands) {
    bool flag = true;
    for (int i = 0; i < c.size(); i++) {
      for (int j = i + 1; j < c.size(); j++) {
        int v = c[i], u = c[j];
        if (g[v].find(u) == g[v].end()) {
          flag = false;
          break;
        }
      }
      if (!flag) break;
    }
    if (flag) {
      cout << 2 << "\n";
      for (const int &v : c) {
        cout << (v + 1) << " ";
      }
      cout << "\n";
      return;
    }
  }
  cout << -1 << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  chrono::steady_clock::time_point begin = chrono::steady_clock::now();
  int tests = 1;
  cin >> tests;
  while (tests--) {
    solve();
  }
  chrono::steady_clock::time_point end = chrono::steady_clock::now();
  return 0;
}