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;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n, m, k;
    cin >> n >> m >> k;
    vector<vector<int>> V(n);
    for (int i = 0; i < m; ++i) {
      int u, v;
      cin >> u >> v;
      --v, --u;
      V[u].push_back(v);
      V[v].push_back(u);
    }
    vector<int> que, d(n), in(n, 2);
    for (int i = 0; i < n; ++i) {
      d[i] = V[i].size();
      sort(V[i].begin(), V[i].end());
      if (d[i] < k) {
        que.push_back(i);
        in[i] = 1;
      }
    }
    bool output = 0;
    for (int i = 0; i < que.size(); ++i) {
      if (output) break;
      int u = que[i];
      if (d[u] == k - 1 && (1ll * k * (k - 1) / 2 <= m)) {
        vector<int> q;
        q.push_back(u);
        for (int j : V[u])
          if (in[j] != 0) q.push_back(j);
        assert(q.size() == k);
        bool ok = 1;
        for (int x = 0; x < q.size() && ok; ++x) {
          for (int y = x + 1; y < q.size() && ok; ++y)
            if (x != y) {
              auto it = lower_bound(V[q[x]].begin(), V[q[x]].end(), q[y]);
              if (it == V[q[x]].end() || *it != q[y]) {
                ok = 0;
              }
            }
        }
        if (ok) {
          cout << 2 << "\n";
          for (int j : q) cout << j + 1 << " ";
          cout << "\n";
          output = 1;
        }
      }
      in[u] = 0;
      for (int v : V[u]) {
        d[v]--;
        if (d[v] < k && in[v] == 2) {
          que.push_back(v);
          in[v] = 1;
        }
      }
    }
    if (output) continue;
    if (que.size() != n) {
      cout << 1 << " " << n - que.size() << "\n";
      for (int i = 0; i < n; ++i)
        if (in[i] == 2) {
          cout << i + 1 << " ";
        }
      cout << '\n';
    } else {
      cout << -1 << '\n';
    }
  }
  return 0;
}