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.
#include <bits/stdc++.h>
using namespace std;
using P = pair<int, int>;
const int M = 1000000007;
const long long LM = 1LL << 60;
vector<int> solve(const vector<vector<int>>& edge, int k) {
  int n = edge.size();
  queue<int> q;
  vector<int> deg(n);
  vector<bool> del(n);
  for (int i = 0; i < n; ++i) {
    deg[i] = edge[i].size();
    if (deg[i] < k) {
      del[i] = true;
      q.push(i);
    }
  }
  while (!q.empty()) {
    int p = q.front();
    q.pop();
    for (auto& i : edge[p]) {
      if (del[i]) continue;
      --deg[i];
      if (deg[i] < k) {
        del[i] = true;
        q.push(i);
      }
    }
  }
  vector<int> res;
  for (int i = 0; i < n; ++i) {
    if (!del[i]) {
      res.push_back(i);
    }
  }
  return res;
}
vector<int> solve2(const vector<vector<int>>& edge, int k) {
  int n = edge.size();
  queue<int> q;
  vector<int> deg(n);
  vector<bool> del(n);
  for (int i = 0; i < n; ++i) {
    deg[i] = edge[i].size();
    if (deg[i] < k - 1) {
      del[i] = true;
      q.push(i);
    }
  }
  while (!q.empty()) {
    int p = q.front();
    q.pop();
    for (auto& i : edge[p]) {
      if (del[i]) continue;
      --deg[i];
      if (deg[i] < k - 1) {
        del[i] = true;
        q.push(i);
      }
    }
  }
  vector<bool> pushed(n);
  for (int i = 0; i < n; ++i) {
    if (!del[i] && deg[i] == k - 1) {
      pushed[i] = true;
      q.push(i);
    }
  }
  vector<unordered_set<int>> edgeset(n);
  for (int i = 0; i < n; ++i) {
    for (auto& e : edge[i]) {
      edgeset[i].insert(e);
    }
  }
  while (!q.empty()) {
    int p = q.front();
    q.pop();
    if (deg[p] == k - 1) {
      vector<int> v;
      v.push_back(p);
      for (auto& i : edge[p]) {
        if (!del[i]) {
          v.push_back(i);
        }
      }
      bool ok = true;
      for (int i = 0; i < k && ok; ++i) {
        for (int j = i + 1; j < k; ++j) {
          if (!edgeset[v[i]].count(v[j])) {
            ok = false;
            break;
          }
        }
      }
      if (ok) {
        return v;
      }
    }
    for (auto& i : edge[p]) {
      if (del[i]) continue;
      --deg[i];
      if (deg[i] == k - 1 && !pushed[i]) {
        pushed[i] = true;
        q.push(i);
      }
    }
    del[p] = true;
  }
  return {};
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  int T;
  cin >> T;
  for (int _ = 0; _ < T; ++_) {
    int n, m, k;
    cin >> n >> m >> k;
    vector<vector<int>> edge(n);
    for (int i = 0; i < m; ++i) {
      int u, v;
      cin >> u >> v;
      --u;
      --v;
      edge[u].push_back(v);
      edge[v].push_back(u);
    }
    {
      vector<int> s0 = solve(edge, k);
      if (s0.size() > 0) {
        cout << 1 << ' ' << s0.size() << '\n';
        for (int i = 0; i < (int)s0.size(); ++i) {
          cout << s0[i] + 1 << (i + 1 < (int)s0.size() ? ' ' : '\n');
        }
        continue;
      }
    }
    {
      vector<int> s1 = solve2(edge, k);
      if (s1.size() > 0) {
        cout << 2 << '\n';
        for (int i = 0; i < (int)s1.size(); ++i) {
          cout << s1[i] + 1 << (i + 1 < (int)s1.size() ? ' ' : '\n');
        }
        continue;
      }
    }
    cout << -1 << '\n';
  }
  return 0;
}
