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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <class A, class B>
ostream& operator<<(ostream& out, const pair<A, B>& a) {
  return out << "(" << a.first << ", " << a.second << ")";
}
template <int>
ostream& operator<<(ostream& os, const vector<int>& v) {
  os << "[";
  for (int i = 0; i < v.size(); ++i) {
    if (v[i] != 1000000001)
      os << v[i];
    else
      os << "INF";
    if (i != v.size() - 1) os << ", ";
  }
  os << "]\n";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "[";
  for (int i = 0; i < v.size(); ++i) {
    os << v[i];
    ;
    if (i != v.size() - 1) os << ", ";
  }
  os << "]\n";
  return os;
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  unordered_set<int> adj[n], adjEdge[n];
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    adj[a].insert(b);
    adj[b].insert(a);
    adjEdge[a].insert(b);
    adjEdge[b].insert(a);
  }
  set<pair<int, int> > degVertex;
  int vertexDegMap[n];
  set<int> vertices;
  for (int i = 0; i < n; i++) {
    vertices.insert(i);
    vertexDegMap[i] = adj[i].size();
    degVertex.insert(make_pair(adj[i].size(), i));
  }
  if (1LL * k * (k - 1) > 2 * m) {
    cout << -1 << endl;
    return;
  }
  vector<int> candidates[n];
  while (!degVertex.empty() && (*degVertex.begin()).first < k) {
    pair<int, int> p = *degVertex.begin();
    int vert = p.second, deg = p.first;
    if (deg == k - 1) {
      for (auto v : adj[vert]) {
        if (!vertices.count(v)) continue;
        candidates[vert].push_back(v);
      }
      candidates[vert].push_back(vert);
    }
    for (auto v : adj[vert]) {
      if (!vertices.count(v)) continue;
      int degV = vertexDegMap[v];
      degVertex.erase(make_pair(degV, v));
      vertexDegMap[v]--;
      adj[v].erase(vert);
      if (vertexDegMap[v] != 0) {
        degVertex.insert(make_pair(degV - 1, v));
      } else
        vertices.erase(v);
    }
    degVertex.erase(p);
    vertices.erase(vert);
    adj[vert].clear();
  }
  if (vertices.empty()) {
    for (int i = 0; i < n; i++) {
      if (candidates[i].size() != k) continue;
      int vert = i;
      bool flag = true;
      for (auto v : candidates[vert]) {
        for (auto w : candidates[vert]) {
          if (v == w) continue;
          if ((adjEdge[v].find(w) == adjEdge[v].end())) {
            flag = false;
            break;
          }
        }
        if (!flag) break;
      }
      if (flag) {
        cout << 2 << endl;
        for (auto v : candidates[vert]) cout << v + 1 << " ";
        cout << endl;
        return;
      }
    }
    cout << -1 << endl;
  } else {
    cout << 1 << " " << vertices.size() << endl;
    for (auto v : vertices) cout << v + 1 << " ";
    cout << endl;
  }
}
int main() {
  {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
  };
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}