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>
#pragma GCC optimize("Ofast")
using namespace std;
vector<unordered_set<int>> g;
vector<unordered_set<int>> g2;
vector<int> deg;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long T = 1, n, m, k;
  cin >> T;
  while (T--) {
    cin >> n >> m >> k;
    g.clear();
    g2.clear();
    g.assign(n, unordered_set<int>());
    g2.assign(n, unordered_set<int>());
    deg.assign(n, 0);
    for (int i = 0; i < m; ++i) {
      int u, v;
      cin >> u >> v;
      u--, v--;
      deg[u]++;
      deg[v]++;
      g[u].insert(v);
      g2[u].insert(v);
      g[v].insert(u);
      g2[v].insert(u);
    }
    if ((long long)k * (k - 1ll) / 2ll > m) {
      cout << "-1\n";
      continue;
    }
    set<pair<int, int>> active;
    for (int j = 0; j < n; ++j) {
      active.insert({deg[j], j});
    }
    int ans = 0;
    vector<unordered_set<int>> cliques;
    while (active.size() >= k) {
      pair<int, int> p = *active.begin();
      int u = p.second;
      if (p.first >= k) {
        ans = 1;
        break;
      }
      if (p.first == k - 1) {
        cliques.push_back(unordered_set<int>());
        cliques[cliques.size() - 1].insert(u);
        for (int v : g[u]) cliques[cliques.size() - 1].insert(v);
      }
      active.erase(p);
      for (int v : g[u]) {
        active.erase({deg[v], v});
        deg[v]--;
        active.insert({deg[v], v});
        g[v].erase(u);
      }
    }
    if (ans == 1) {
      cout << "1 " << active.size() << "\n";
      for (pair<int, int> a : active) cout << a.second + 1 << " ";
      cout << "\n";
      continue;
    }
    int isClique = 0;
    for (int l = 0; l < cliques.size(); ++l) {
      isClique = l + 1;
      for (int u : cliques[l]) {
        for (int v : cliques[l]) {
          if (u == v) continue;
          if (g2[u].find(v) == g2[u].end()) isClique = 0;
          if (!isClique) break;
        }
        if (!isClique) break;
      }
      if (isClique) break;
    }
    if (isClique) {
      cout << "2\n";
      for (int u : cliques[isClique - 1]) cout << u + 1 << " ";
      cout << "\n";
      continue;
    }
    cout << "-1\n";
  }
}