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;
constexpr int maxn = 100005, maxm = 100005;
unordered_set<int> adj[maxn];
bool iq2[maxn], removed[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n, m, k;
    cin >> n >> m >> k;
    for (int i = 0; i < n; i++) {
      adj[i].clear();
    }
    for (int i = 0; i < m; i++) {
      int u, v;
      cin >> u >> v;
      u--;
      v--;
      adj[u].insert(v);
      adj[v].insert(u);
    }
    if ((long long)k * (k - 1) / 2 > m) {
      cout << "-1\n";
      continue;
    }
    deque<int> q2;
    fill_n(iq2, n, false);
    fill_n(removed, n, false);
    for (int i = 0; i < n; i++) {
      if ((int)adj[i].size() < k - 1) {
        q2.push_front(i);
        iq2[i] = true;
      } else if ((int)adj[i].size() == k - 1) {
        q2.push_back(i);
        iq2[i] = true;
      }
    }
    bool good = false;
    while (!q2.empty()) {
      if ((long long)k * (k - 1) / 2 > m) {
        cout << "-1\n";
        good = true;
        break;
      }
      int u = q2.front();
      q2.pop_front();
      if ((int)adj[u].size() == k - 1) {
        for (auto it = adj[u].begin(); it != adj[u].end(); ++it) {
          for (auto jt = next(it); jt != adj[u].end(); ++jt) {
            if (!adj[*it].count(*jt)) {
              goto l1;
            }
          }
        }
        cout << "2\n";
        cout << u + 1 << ' ';
        for (int v : adj[u]) {
          cout << v + 1 << ' ';
        }
        cout << '\n';
        good = true;
        break;
      }
    l1:
      removed[u] = true;
      m -= (int)adj[u].size();
      for (int v : adj[u]) {
        adj[v].erase(u);
        if (!iq2[v]) {
          if ((int)adj[v].size() < k - 1) {
            q2.push_front(v);
            iq2[v] = true;
          } else if ((int)adj[v].size() == k - 1) {
            q2.push_back(v);
            iq2[v] = true;
          }
        }
      }
    }
    if (!good) {
      long int cnt = count(removed, removed + n, false);
      if (cnt > 0) {
        cout << "1 " << cnt << '\n';
        for (int i = 0; i < n; i++) {
          if (!removed[i]) {
            cout << i + 1 << ' ';
          }
        }
        cout << '\n';
      } else {
        cout << "-1\n";
      }
    }
  }
}