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>
std::mt19937 rng(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
void solve() {
  int n, m, k;
  std::cin >> n >> m >> k;
  std::vector<std::vector<int>> edges(n);
  std::vector<int> deg(n, 0);
  for (int i = 0; i < m; i++) {
    int u, v;
    std::cin >> u >> v;
    u--;
    v--;
    deg[u]++;
    deg[v]++;
    edges[u].push_back(v);
    edges[v].push_back(u);
  }
  std::queue<int> q;
  for (int i = 0; i < n; i++) {
    if (deg[i] < k) {
      q.push(i);
    }
    std::sort(edges[i].begin(), edges[i].end());
  }
  std::vector<bool> good(n, false), dirty(n, false);
  std::vector<std::vector<int>> haha(n);
  while (!q.empty()) {
    int on = q.front();
    q.pop();
    dirty[on] = true;
    for (auto to : edges[on]) {
      if (deg[to] >= k - 1 && !dirty[to]) {
        haha[on].push_back(to);
      }
      deg[to]--;
      if (deg[to] == k - 1) {
        q.push(to);
      }
    }
    good[on] = (int)haha[on].size() == k - 1 && deg[on] == k - 1;
  }
  std::vector<int> ans;
  for (int i = 0; i < n; i++) {
    if (deg[i] >= k) {
      ans.push_back(i);
    }
  }
  if (!ans.empty()) {
    std::cout << "1 " << ans.size() << "\n";
    for (int i = 0; i < (int)ans.size(); i++) {
      std::cout << ans[i] + 1 << (i + 1 == (int)ans.size() ? '\n' : ' ');
    }
    return;
  }
  auto ask = [&](int u, int v) {
    auto it = std::lower_bound(edges[u].begin(), edges[u].end(), v);
    return it != edges[u].end() && *it == v;
  };
  for (int i = 0; i < n && (long long)k * (k - 1) / 2 <= m; i++) {
    for (int j = (int)haha[i].size() - 2; j >= 0 && good[i]; j--) {
      for (int jj = j + 1; jj < (int)haha[i].size() && good[i]; jj++) {
        good[i] = good[i] && ask(haha[i][j], haha[i][jj]);
      }
    }
    if (good[i]) {
      ans = haha[i];
      ans.push_back(i);
      break;
    }
  }
  if (!ans.empty()) {
    std::cout << "2\n";
    for (int i = 0; i < (int)ans.size(); i++) {
      std::cout << ans[i] + 1 << (i + 1 == (int)ans.size() ? '\n' : ' ');
    }
    return;
  }
  std::cout << "-1\n";
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  int t;
  std::cin >> t;
  while (t--) {
    solve();
  }
}