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;
const int inf = 1e9;
const int mod = 1e9 + 7;
const int p = 1557;
void solve() {
  int n, m, k, u, v;
  cin >> n >> m >> k;
  vector<unordered_set<int>> adj(n + 1);
  for (int i = 0; i < m; i++) {
    cin >> u >> v;
    adj[u].insert(v);
    adj[v].insert(u);
  }
  priority_queue<array<int, 2>> pq;
  for (int i = 1; i <= n; i++) {
    pq.push({-(int)adj[i].size(), i});
  }
  while (pq.size()) {
    int node = pq.top()[1], deg = -pq.top()[0];
    pq.pop();
    if (adj[node].size() != deg) continue;
    if (deg >= k) {
      vector<int> store;
      for (int i = 1; i <= n; i++) {
        if (adj[i].size() >= k) {
          store.push_back(i);
        }
      }
      cout << 1 << " " << store.size() << "\n";
      for (int temp : store) {
        cout << temp << " ";
      }
      cout << "\n";
      return;
    }
    vector<int> neigh;
    for (int nxt : adj[node]) {
      neigh.push_back(nxt);
    }
    if (deg < k - 1) {
      for (int nxt : neigh) {
        adj[nxt].erase(node);
        adj[node].erase(nxt);
        pq.push({-(int)adj[nxt].size(), nxt});
      }
    } else {
      int ok = 1;
      for (int i = 0; i < neigh.size(); i++) {
        for (int j = i + 1; j < neigh.size(); j++) {
          if (!adj[neigh[i]].count(neigh[j])) {
            ok = 0;
            break;
          }
        }
        if (!ok) break;
      }
      if (ok) {
        cout << 2 << "\n";
        for (int nxt : neigh) {
          cout << nxt << " ";
        }
        cout << node << " \n";
        return;
      } else {
        for (int nxt : neigh) {
          adj[nxt].erase(node);
          adj[node].erase(nxt);
          pq.push({-(int)adj[nxt].size(), nxt});
        }
      }
    }
  }
  cout << -1 << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  std::cout << std::fixed << std::setprecision(20);
  ;
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  }
}