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 rr[] = {0, 1, 1, 1, 0, -1, -1, -1};
int cc[] = {1, 1, 0, -1, -1, -1, 0, 1};
const int mod = 1e9 + 7;
const int N = 300050;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
vector<unordered_map<int, bool, custom_hash>> edges;
vector<bool> alive;
vector<int> degree;
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  degree.assign(n, 0);
  edges.assign(n, unordered_map<int, bool, custom_hash>());
  alive.assign(n, true);
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    edges[a][b];
    edges[b][a];
  }
  if (k > 500) {
    cout << -1 << '\n';
    return;
  }
  queue<int> Q;
  for (int i = 0; i < n; i++) {
    degree[i] = edges[i].size();
    if (alive[i] && degree[i] < k) {
      Q.push(i);
    }
  }
  while ((int)Q.size()) {
    int cur = Q.front();
    Q.pop();
    if (degree[cur] == k - 1) {
      vector<int> clique;
      clique.push_back(cur);
      for (auto &x : edges[cur]) {
        if (alive[x.first]) {
          clique.push_back(x.first);
        }
      }
      bool poss = true;
      for (int i = 0; i < k && poss; i++) {
        for (int j = i + 1; j < k && poss; j++) {
          if (!edges[clique[i]].count(clique[j])) {
            poss = false;
          }
        }
      }
      if (poss) {
        cout << "2\n";
        for (auto &x : clique) {
          cout << x + 1 << ' ';
        }
        cout << '\n';
        return;
      } else {
        alive[cur] = false;
        for (int i = 1; i < k; i++) {
          if (degree[clique[i]] == k) {
            Q.push(clique[i]);
          }
          degree[clique[i]]--;
        }
      }
    } else {
      alive[cur] = false;
      for (auto &x : edges[cur]) {
        if (!alive[x.first]) {
          continue;
        }
        if (degree[x.first] == k) {
          Q.push(x.first);
        }
        degree[x.first]--;
      }
    }
  }
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    if (alive[i]) {
      ans.push_back(i);
    }
  }
  if ((int)ans.size() > 0) {
    cout << 1 << ' ' << ans.size() << '\n';
    for (auto &x : ans) {
      cout << x + 1 << ' ';
    }
    cout << '\n';
  } else {
    cout << "-1\n";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int test = 1;
  cin >> test;
  while (test--) {
    solve();
  }
  return 0;
}