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("O3,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
mt19937 rnd(time(nullptr));
const int N = 1e5 + 5;
unordered_set<int> g[N];
bool check(int v) {
  for (auto i : g[v])
    for (auto j : g[v])
      if (i != j && !g[i].count(j)) return false;
  cout << "2\n" << v << " ";
  for (auto i : g[v]) cout << i << " ";
  cout << "\n";
  return true;
}
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 1; i <= n; ++i) g[i].clear();
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].insert(v), g[v].insert(u);
  }
  set<pair<int, int> > st;
  for (int i = 1; i <= n; ++i) {
    st.insert({g[i].size(), i});
  }
  if ((long long)k * (k - 1) > 2 * m) {
    cout << "-1\n";
    return;
  }
  while (st.size()) {
    pair<int, int> v = *st.begin();
    if (v.first >= k) break;
    if (v.first == k - 1 && check(v.second)) return;
    st.erase(st.begin());
    for (auto to : g[v.second]) {
      st.erase({g[to].size(), to});
      g[to].erase(v.second);
      st.insert({g[to].size(), to});
    }
  }
  if (st.size()) {
    cout << "1 " << st.size() << "\n";
    for (auto v : st) cout << v.second << " ";
    cout << "\n";
  } else
    cout << "-1\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}