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>
#pragma GCC optimize(2)
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cout.tie(0);
  int T = 1;
  cin >> T;
  while (T--) {
    int n, m, k;
    cin >> n >> m >> k;
    vector<vector<int>> e(n);
    for (int i = 0; i < m; i++) {
      int u, v;
      cin >> u >> v;
      --u, --v;
      e[u].push_back(v);
      e[v].push_back(u);
    }
    vector<int> deg(n);
    for (int i = 0; i < n; i++)
      sort(e[i].begin(), e[i].end()), deg[i] = e[i].size();
    queue<int> q;
    vector<int> vis(n);
    for (int i = 0; i < n; i++)
      if (deg[i] < k) vis[i] = 1, q.push(i);
    vector<int> clique;
    while (q.size()) {
      int u = q.front();
      q.pop();
      vis[u] = 2;
      if (deg[u] == k - 1 && clique.empty() && 1ll * k * (k - 1) / 2 <= m) {
        clique.push_back(u);
        for (auto v : e[u])
          if (vis[v] < 2) clique.push_back(v);
        bool ok = 1;
        for (auto x : clique)
          for (auto y : clique)
            if (x != y && !binary_search(e[x].begin(), e[x].end(), y)) ok = 0;
        if (!ok) clique.clear();
      }
      for (auto v : e[u])
        if (--deg[v] < k && !vis[v]) vis[v] = 1, q.push(v);
    }
    int cnt = 0;
    for (int i = 0; i < n; i++) cnt += vis[i] == 0;
    if (cnt > 0) {
      cout << 1 << ' ' << cnt << '\n';
      for (int i = 0; i < n; i++)
        if (vis[i] == 0) cout << i + 1 << ' ';
      cout << '\n';
    } else if (clique.size()) {
      cout << 2 << '\n';
      for (auto t : clique) cout << t + 1 << ' ';
      cout << '\n';
    } else
      cout << -1 << '\n';
  }
  return 0;
}