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 double eps = 1e-8;
const long long mod = 998244353;
const int inf = 1 << 30;
const int maxn = 1 << 21;
int n, m;
unordered_set<int> adj[maxn];
int deg[maxn];
bool vs[maxn];
vector<int> getClique(int u) {
  vector<int> cc = {u};
  for (auto v : adj[u])
    if (!vs[v]) cc.push_back(v);
  for (int i = (1); i < (int)(cc.size()); ++i)
    for (int j = (i + 1); j < (int)(cc.size()); ++j)
      if (!adj[cc[i]].count(cc[j])) return {};
  return cc;
}
void solve() {
  int k;
  cin >> n >> m >> k;
  for (int i = 0; i < (int)(n + 1); ++i) adj[i].clear(), vs[i] = 0, deg[i] = 0;
  for (int i = 0; i < (int)(m); ++i) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    adj[u].insert(v);
    adj[v].insert(u);
    deg[u]++;
    deg[v]++;
  }
  set<pair<int, int>> ss;
  for (int i = 0; i < (int)(n); ++i) ss.insert({deg[i], i});
  while (!ss.empty()) {
    int u = ss.begin()->second;
    if (deg[u] >= k) {
      cout << "1 " << ss.size() << "\n";
      for (auto [_, v] : ss) cout << v + 1 << " ";
      cout << "\n";
      return;
    }
    if (deg[u] + 1 == k) {
      vector<int> cc = getClique(u);
      if (!cc.empty()) {
        cout << "2\n";
        for (auto v : cc) cout << v + 1 << " ";
        cout << "\n";
        return;
      }
    }
    vs[u] = 1;
    ss.erase(ss.begin());
    for (auto v : adj[u])
      if (!vs[v]) {
        ss.erase({deg[v], v});
        ss.insert({--deg[v], v});
      }
  }
  cout << -1 << "\n";
}
int main() {
  int tc = 1;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> tc;
  for (int tt = 0; tt < (int)(tc); ++tt) {
    solve();
  }
}