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 int MXN = 1e5 + 5;
int n, m, k;
vector<int> adj[MXN];
int du[MXN], del[MXN];
int cnt[MXN];
void work() {
  cin >> n >> m >> k;
  for (register int i = (0), LIM = (m); i < LIM; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].emplace_back(v), adj[v].emplace_back(u);
    ++du[u], ++du[v];
  }
  set<pair<int, int> > Q;
  for (register int i = (1), LIM = (n + 1); i < LIM; ++i) {
    Q.insert(make_pair(du[i], i));
    sort(adj[i].begin(), adj[i].end());
  }
  int flag = 0;
  vector<int> vs;
  while (!Q.empty() && flag == 0) {
    pair<int, int> u = *Q.begin();
    if (u.first == k - 1) {
      vs.clear();
      for (int v : adj[u.second]) {
        if (del[v] == 0) {
          vs.emplace_back(v);
        }
      }
      vs.emplace_back(u.second);
      flag = 2;
      for (int a : vs) {
        if (a == u.second) break;
        for (int b : vs) {
          if (a != b) {
            if (!binary_search(adj[a].begin(), adj[a].end(), b)) {
              flag = 0;
              break;
            }
          }
        }
        if (flag == 0) break;
      }
      if (flag == 0) {
        auto p = Q.lower_bound(make_pair(du[u.second], u.second));
        Q.erase(p);
        del[u.second] = 1;
        for (int v : adj[u.second]) {
          if (del[v]) continue;
          --du[v];
          auto p = Q.lower_bound(make_pair(du[v] + 1, v));
          Q.erase(p);
          Q.insert(make_pair(du[v], v));
        }
      }
    } else if (u.first >= k) {
      flag = 1;
    } else {
      auto p = Q.lower_bound(make_pair(du[u.second], u.second));
      Q.erase(p);
      del[u.second] = 1;
      for (int v : adj[u.second]) {
        if (del[v]) continue;
        --du[v];
        auto p = Q.lower_bound(make_pair(du[v] + 1, v));
        Q.erase(p);
        Q.insert(make_pair(du[v], v));
      }
    }
  }
  if (flag == 0)
    printf("-1\n");
  else if (flag == 1) {
    printf("1 %d\n", Q.size());
    for (pair<int, int> x : Q) printf("%d ", x.second);
    printf("\n");
  } else {
    printf("2\n");
    for (int x : vs) printf("%d ", x);
    printf("\n");
  }
  for (register int i = (1), LIM = (n + 1); i < LIM; ++i) {
    adj[i].clear();
    du[i] = del[i] = 0;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tim = 1;
  cin >> tim;
  for (int cas = 1; cas <= tim; ++cas) work();
  return 0;
}