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;
using ll = long long;
const int NAX = 2e5 + 5, MOD = 1000000007;
void solveCase() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<vector<int>> adj(n);
  vector<vector<pair<int, int>>> ch(n);
  vector<int> degree(n);
  vector<int> mark(n);
  vector<int> can(n);
  vector<int> ans(n);
  vector<int> ct(n), A;
  for (size_t i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    degree[u]++;
    degree[v]++;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  if (k * (k - 1LL) > 2 * m) {
    cout << -1 << '\n';
    return;
  }
  set<pair<int, int>> Q;
  for (size_t i = 0; i < n; i++) Q.insert({degree[i], i});
  int cnt = 1;
  while (Q.size()) {
    auto it = *Q.begin();
    if (it.first >= k) {
      cout << 1 << '\n';
      cout << Q.size() << '\n';
      for (auto &x : Q) cout << x.second + 1 << ' ';
      cout << '\n';
      return;
    } else {
      Q.erase(it);
      mark[it.second] = cnt++;
      A.push_back(it.second);
      for (auto &u : adj[it.second])
        if (!mark[u]) {
          Q.erase({degree[u], u});
          Q.insert({--degree[u], u});
        }
    }
  }
  for (auto &i : A) {
    int nxt = 0;
    ;
    for (auto &u : adj[i])
      if (mark[u] > mark[i]) ct[nxt++] = u, can[u] = 1;
    for (auto &u : ch[i])
      if (!can[u.first]) ans[u.second] = false;
    for (size_t j = 0; j < nxt; j++) can[ct[j]] = false;
    if (nxt != k - 1) continue;
    ans[i] = true;
    sort(ct.begin(), ct.begin() + nxt,
         [&](int u, int v) -> bool { return mark[u] < mark[v]; });
    for (size_t j = 0; j < nxt; j++)
      for (size_t k = j + 1; k < nxt; k++) ch[ct[j]].push_back({ct[k], i});
  }
  for (auto &i : A) {
    if (!ans[i]) continue;
    cout << 2 << '\n';
    cout << i + 1 << ' ';
    for (auto &v : adj[i])
      if (mark[v] > mark[i]) cout << v + 1 << ' ';
    cout << '\n';
    return;
  }
  cout << -1 << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; ++i) solveCase();
  return 0;
}