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;
using ld = long double;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
using vint = vector<int>;
using vll = vector<ll>;
using vld = vector<ld>;
using vpii = vector<pii>;
using vpil = vector<pil>;
using vpli = vector<pli>;
using vpll = vector<pll>;
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<vint> e(n + 1);
  vector<pii> elist;
  for (int i = 0; i < m; i++) {
    int first, second;
    cin >> first >> second;
    e[first].push_back(second);
    e[second].push_back(first);
    elist.emplace_back(first, second);
    elist.emplace_back(second, first);
  }
  sort(elist.begin(), elist.end());
  queue<int> q;
  vint deg(n + 1), chk(n + 1);
  for (int i = 1; i <= n; i++) {
    deg[i] = e[i].size();
    if (deg[i] < k) {
      chk[i] = 1;
      q.push(i);
    }
  }
  while (!q.empty()) {
    int first = q.front();
    q.pop();
    chk[first] = 2;
    if (deg[first] == k - 1) {
      vint v = {first};
      for (int i : e[first])
        if (chk[i] < 2) v.push_back(i);
      if (v.size() == k) {
        int is_clique = 1;
        for (int i = 0; i < v.size(); i++) {
          for (int j = i + 1; j < v.size(); j++) {
            if (!binary_search(elist.begin(), elist.end(), pii(v[i], v[j]))) {
              is_clique = 0;
              break;
            }
          }
          if (!is_clique) break;
        }
        if (is_clique) {
          cout << "2\n";
          for (int i : v) cout << i << ' ';
          cout << '\n';
          return;
        }
      }
    }
    for (int i : e[first]) {
      if (chk[i]) continue;
      deg[i]--;
      if (deg[i] < k) {
        chk[i] = 1;
        q.push(i);
      }
    }
  }
  if (count(1 + chk.begin(), chk.end(), 0)) {
    vint ans;
    for (int i = 1; i <= n; i++)
      if (!chk[i]) ans.push_back(i);
    cout << "1 " << ans.size() << '\n';
    for (int i : ans) cout << i << ' ';
    cout << "\n";
  } else
    cout << "-1\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int tc;
  cin >> tc;
  while (tc--) solve();
  return 0;
}