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;
vector<bool> remove_under(const vector<vector<int>>& adj, int k,
                          vector<int>& n_adj) {
  queue<int> q;
  int n = adj.size();
  vector<bool> removed(n, false);
  for (int i = 0; i < n; ++i) {
    if ((int)adj[i].size() < k) {
      removed[i] = true;
      q.push(i);
    }
  }
  while (!q.empty()) {
    int i = q.front();
    q.pop();
    for (int j : adj[i]) {
      if (!removed[j]) {
        n_adj[j]--;
        if (n_adj[j] < k) {
          removed[j] = true;
          q.push(j);
        }
      }
    }
  }
  return removed;
}
void test_case() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> u(m), v(m);
  for (int i = 0; i < m; ++i) {
    cin >> u[i] >> v[i];
    u[i]--;
    v[i]--;
  }
  if (k == 1) {
    cout << 2 << endl << 1 << endl;
    return;
  }
  if ((2 * m) < (k * (k - 1))) {
    cout << -1 << endl;
    return;
  }
  vector<unordered_set<int>> adj(n);
  for (int i = 0; i < m; ++i) {
    adj[u[i]].insert(v[i]);
    adj[v[i]].insert(u[i]);
  }
  priority_queue<pair<int, int>> q;
  vector<int> degrees(n);
  for (int i = 0; i < n; ++i) {
    degrees[i] = adj[i].size();
    q.emplace(-degrees[i], i);
  }
  vector<bool> visited(n, false);
  while (!q.empty()) {
    int degree, me;
    tie(degree, me) = q.top();
    q.pop();
    if (!visited[me]) {
      visited[me] = true;
      degree = -degree;
      if (degree < (k - 1)) {
        for (int other : adj[me]) {
          adj[other].erase(me);
          degrees[other]--;
          q.emplace(-degrees[other], other);
        }
      } else if (degree >= k) {
        vector<int> subset;
        for (int i = 0; i < n; ++i) {
          if (!visited[i]) {
            subset.push_back(i);
          }
        }
        int s = subset.size();
        cout << 1 << " " << (s + 1) << endl;
        cout << (me + 1);
        for (int i = 0; i < s; ++i) {
          cout << " " << (subset[i] + 1);
        }
        cout << endl;
        return;
      } else {
        vector<int> clique(k);
        clique[0] = me;
        int idx = 1;
        for (int other : adj[me]) {
          clique[idx++] = other;
        }
        assert(idx == k);
        bool is_clique = true;
        for (int i = 1; is_clique && (i < k); ++i) {
          for (int j = 0; is_clique && (j < k); ++j) {
            if ((j != i) && (adj[clique[j]].count(clique[i]) == 0)) {
              is_clique = false;
            }
          }
        }
        if (is_clique) {
          cout << 2 << endl;
          cout << (clique[0] + 1);
          for (int i = 1; i < k; ++i) {
            cout << " " << (clique[i] + 1);
          }
          cout << endl;
          return;
        }
        for (int other : adj[me]) {
          adj[other].erase(me);
          degrees[other]--;
          q.emplace(-degrees[other], other);
        }
      }
    }
  }
  cout << -1 << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  for (int i = 0; i < t; ++i) {
    test_case();
  }
}