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>
const int N = 100005;
int n, m, K;
long long hash[N << 1];
std::vector<int> E[N];
int d[N];
bool check(const std::vector<int> &id) {
  for (int i = 0; i < (int)id.size(); ++i) {
    for (int j = i + 1; j < (int)id.size(); ++j) {
      long long t = 1ll * id[i] * n + id[j];
      int p = std::lower_bound(hash + 1, hash + 1 + 2 * m, t) - hash;
      if (p > 2 * m || hash[p] != t) {
        return false;
      }
    }
  }
  return true;
}
void solve() {
  std::cin >> n >> m >> K;
  for (int i = 1; i <= n; ++i) {
    d[i] = 0;
    E[i].clear();
  }
  for (int i = 1; i <= m; ++i) {
    int u, v;
    std::cin >> u >> v;
    hash[i] = 1ll * u * n + v;
    hash[i + m] = 1ll * v * n + u;
    ++d[u], ++d[v];
    E[u].push_back(v);
    E[v].push_back(u);
  }
  if (1ll * K * (K - 1) / 2 > m) {
    std::cout << -1 << "\n";
    return;
  }
  std::sort(hash + 1, hash + 1 + 2 * m);
  std::vector<int> S;
  for (int i = 1; i <= n; ++i) {
    if (d[i] < K) {
      S.push_back(i);
    }
  }
  while (!S.empty()) {
    int u = S.back();
    if (d[u] == K - 1) {
      std::vector<int> id;
      id.push_back(u);
      for (int v : E[u]) {
        if (~d[v] && d[v] >= K - 1) {
          id.push_back(v);
        }
      }
      if ((int)id.size() == K && check(id)) {
        std::cout << 2 << "\n";
        for (int v : id) {
          std::cout << v << " ";
        }
        std::cout << "\n";
        return;
      }
    }
    S.pop_back();
    for (int v : E[u]) {
      if (~d[v]) {
        if (d[v] == K) {
          S.push_back(v);
        }
        --d[v];
      }
    }
    d[u] = -1;
  }
  std::vector<int> ans;
  for (int i = 1; i <= n; ++i) {
    if (~d[i]) {
      ans.push_back(i);
    }
  }
  if (!ans.empty()) {
    std::cout << 1 << " " << ans.size() << "\n";
    for (auto v : ans) {
      std::cout << v << " ";
    }
    std::cout << "\n";
    return;
  }
  std::cout << -1 << "\n";
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(0);
  int T = 1;
  std::cin >> T;
  while (T--) {
    solve();
  }
}