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;
const long long MOD = 1e9 + 7, Base = 998244353;
const long long N = 6e5 + 7;
const long long INF = 1LL * 1000 * 1000 * 1000 * 1000 * 1000 * 1000 + 7LL,
                INF2 = (1LL << 62);
const double pie = acos(-1.0);
const double EPS = 1e-9;
long long n, test, m, u, v, k, Need[N];
vector<long long> e[N];
set<long long> adj[N];
bool Clique(vector<long long> v, long long nod) {
  v.push_back(nod);
  for (long long u : v) Need[u] = 1;
  long long sz = (long long)v.size(), tot;
  tot = (sz * (sz - 1)) / 2;
  for (long long par : v) {
    for (long long child : e[par]) {
      if (Need[child]) {
        --tot;
      }
    }
  }
  for (long long u : v) Need[u] = 0;
  if (tot) return false;
  cout << 2 << '\n';
  for (long long u : v) cout << u << ' ';
  cout << '\n';
  return true;
}
void Clear() {
  for (long long i = 1; i <= n; ++i) {
    if (adj[i].size()) adj[i].clear();
    while (e[i].size()) e[i].pop_back();
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> test;
  while (test--) {
    cin >> n >> m >> k;
    for (long long i = 1; i <= m; ++i)
      cin >> u >> v, adj[u].insert(v), adj[v].insert(u);
    set<pair<long long, long long>> Deg;
    for (long long i = 1; i <= n; ++i) {
      Deg.insert({(long long)adj[i].size(), i});
    }
    while (Deg.size() and (*Deg.begin()).first <= k - 1) {
      u = (*Deg.begin()).second;
      for (long long v : adj[u]) {
        e[u].push_back(v);
        Deg.erase(Deg.find({adj[v].size(), v}));
        adj[v].erase(adj[v].find(u));
        Deg.insert({adj[v].size(), v});
      }
      Deg.erase(Deg.find({(long long)adj[u].size(), u}));
    }
    if (Deg.size()) {
      Clear();
      cout << 1 << ' ' << (long long)Deg.size() << '\n';
      for (pair<long long, long long> it : Deg) cout << it.second << ' ';
      cout << '\n';
      continue;
    }
    bool Poss = false;
    for (long long i = 1; i <= n and !Poss; ++i) {
      if (e[i].size() == k - 1) {
        if (Clique(e[i], i)) {
          Poss = true;
        }
      }
    }
    if (!Poss) cout << -1 << '\n';
    Clear();
  }
  return 0;
}