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 nax = 3e5 + 10;
int n, m, k;
vector<int> g[nax];
vector<int> new_g[nax];
vector<pair<int, int>> edges;
int deg[nax];
bool del[nax];
bool vis[nax];
vector<pair<int, int>> connect;
void solve() {
  cin >> n >> m >> k;
  for (int i = 1; i <= m; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
    edges.emplace_back(u, v);
    ++deg[u];
    ++deg[v];
    connect.push_back(minmax(u, v));
  }
  sort(connect.begin(), connect.end());
  set<pair<int, int>> setik;
  for (int i = 1; i <= n; ++i) {
    setik.emplace(deg[i], i);
  }
  auto my_erase = [&](int x, int y) -> void { setik.erase(make_pair(x, y)); };
  while (!setik.empty() && setik.begin()->first < k) {
    auto [u, v] = *setik.begin();
    setik.erase(setik.begin());
    for (int to : g[v]) {
      if (del[to]) continue;
      my_erase(deg[to], to);
      --deg[to];
      if (deg[to] > 0)
        setik.emplace(deg[to], to);
      else
        del[to] = true;
    }
    del[v] = true;
  }
  if (!setik.empty()) {
    for (auto& [u, v] : edges) {
      if (del[u] || del[v]) continue;
      new_g[u].push_back(v);
      new_g[v].push_back(u);
    }
    vector<int> ans;
    function<void(int)> dfs = [&](int node) -> void {
      vis[node] = true;
      ans.push_back(node);
      for (int to : new_g[node]) {
        if (!vis[to]) dfs(to);
      }
    };
    for (int i = 1; i <= n; ++i) {
      if (!del[i]) {
        dfs(i);
        break;
      }
    }
    cout << "1 " << ans.size() << '\n';
    for (int i : ans) cout << i << ' ';
    cout << '\n';
    return;
  }
  setik.clear();
  for (int i = 1; i <= n; ++i) deg[i] = 0, del[i] = false;
  for (auto& [u, v] : edges) ++deg[u], ++deg[v];
  for (int i = 1; i <= n; ++i) setik.emplace(deg[i], i);
  vector<int> bar;
  while (true) {
    while (!setik.empty() && setik.begin()->first < k - 1) {
      auto [u, v] = *setik.begin();
      setik.erase(setik.begin());
      for (int to : g[v]) {
        if (del[to]) continue;
        my_erase(deg[to], to);
        --deg[to];
        if (deg[to] > 0)
          setik.emplace(deg[to], to);
        else
          del[to] = true;
      }
      del[v] = true;
    }
    if (setik.empty()) break;
    assert(setik.begin()->first == k - 1);
    auto [u, v] = *setik.begin();
    bar.clear();
    for (int to : g[v]) {
      if (!del[to]) bar.push_back(to);
    }
    bar.push_back(v);
    for (int node : bar)
      if (deg[node] < k - 1) {
        goto non_clique;
      }
    for (int i = 0; i < bar.size(); ++i)
      for (int j = i + 1; j < bar.size(); ++j) {
        pair<int, int> foo = minmax(bar[i], bar[j]);
        if (!binary_search(connect.begin(), connect.end(), foo)) {
          goto non_clique;
        }
      }
    cout << "2\n";
    for (int i : bar) cout << i << ' ';
    cout << '\n';
    return;
  non_clique:;
    setik.erase(setik.begin());
    for (int to : g[v]) {
      if (del[to]) continue;
      my_erase(deg[to], to);
      --deg[to];
      if (deg[to] > 0)
        setik.emplace(deg[to], to);
      else
        del[to] = true;
    }
    del[v] = true;
  }
  cout << -1 << '\n';
}
int main() {
  cin.tie(0)->sync_with_stdio(false);
  int T;
  for (cin >> T; T--;) {
    solve();
    for (int i = 1; i <= n; ++i) {
      deg[i] = 0;
      g[i].clear();
      new_g[i].clear();
      del[i] = false;
      vis[i] = false;
    }
    connect.clear();
    edges.clear();
  }
}