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;
class Response {
 public:
  static const int NOTHING = -1;
  static const int SUBSET = 1;
  static const int CLIQUE = 2;

 public:
  int ANSWER_TYPE = NOTHING;
  vector<int> vertices;
  Response() : ANSWER_TYPE(NOTHING) {}
  Response(int typ, vector<int> v) : ANSWER_TYPE(typ), vertices(v) {}
};
class Solver {
 private:
  int n, m, k;
  vector<vector<int>> edge;
  vector<unordered_set<int>> neighbor;
  vector<bool> isErased;
  priority_queue<pair<int, int>> wait;

 private:
  inline bool checkEligibility(long long k, long long m) {
    return k * (k - 1) <= 2 * m;
  }
  void deleteNode(int u) {
    for (int v : neighbor[u]) {
      neighbor[v].erase(u);
      wait.push(make_pair(-neighbor[v].size(), v));
    }
    isErased[u] = true;
  }
  bool checkClique(vector<int> &a) {
    for (int i = 0; i < k; i++) {
      const int u = a[i];
      for (int j = i + 1; j < k; j++) {
        const int v = a[j];
        if (neighbor[u].find(v) == neighbor[u].end()) return false;
      }
    }
    return true;
  }

 public:
  Solver(int n, int m, int k, const vector<pair<int, int>> &edgeList) {
    this->n = n;
    this->m = m;
    this->k = k;
    edge = vector<vector<int>>(n + 1);
    neighbor = vector<unordered_set<int>>(n + 1);
    for (pair<int, int> p : edgeList) {
      edge[p.first].push_back(p.second);
      edge[p.second].push_back(p.first);
      neighbor[p.first].insert(p.second);
      neighbor[p.second].insert(p.first);
    }
  }
  Response findSomething() {
    if (!checkEligibility(k, m)) return Response();
    isErased = vector<bool>(n + 1, false);
    for (int i = 1; i <= n; i++) wait.push(make_pair(-neighbor[i].size(), i));
    while (!wait.empty()) {
      int v = wait.top().second;
      wait.pop();
      if (isErased[v]) continue;
      int deg = neighbor[v].size();
      if (deg >= k) {
        vector<int> alive;
        for (int i = 1; i <= n; i++)
          if (!isErased[i]) alive.push_back(i);
        return Response(Response::SUBSET, alive);
      } else {
        if (deg == k - 1) {
          vector<int> cliqueCandidate;
          cliqueCandidate.push_back(v);
          for (int u : neighbor[v]) cliqueCandidate.push_back(u);
          if (checkClique(cliqueCandidate))
            return Response(Response::CLIQUE, cliqueCandidate);
        }
        deleteNode(v);
      }
    }
    return Response();
  }
};
int main() {
  int tc;
  scanf("%d", &tc);
  while (tc--) {
    int n, m, k;
    scanf("%d%d%d", &n, &m, &k);
    vector<pair<int, int>> edgeList(m);
    for (int i = 0; i < m; i++)
      scanf("%d%d", &edgeList[i].first, &edgeList[i].second);
    Solver bot(n, m, k, edgeList);
    Response ans = bot.findSomething();
    if (ans.ANSWER_TYPE == Response::NOTHING) {
      printf("-1\n");
    } else {
      printf("%d", ans.ANSWER_TYPE);
      if (ans.ANSWER_TYPE == Response::SUBSET)
        printf(" %d", ans.vertices.size());
      printf("\n");
      for (int v : ans.vertices) printf("%d ", v);
      printf("\n");
    }
  }
  return 0;
}