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;
using ld = long double;
using ll = int64_t;
using pii = pair<int, int>;
using ull = uint64_t;
unordered_set<int> g[100100];
int used[100100];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int ttt;
  cin >> ttt;
  for (int tst = 1; tst <= ttt; ++tst) {
    int n, m, k;
    cin >> n >> m >> k;
    for (int i = 1; i <= n; ++i) g[i].clear();
    for (int i = 0; i < m; ++i) {
      int x, y;
      cin >> x >> y;
      g[x].insert(y);
      g[y].insert(x);
    }
    if (k * ll(k - 1) / 2 > m) {
      cout << "-1\n";
      continue;
    }
    priority_queue<pii, vector<pii>, greater<pii>> q;
    for (int i = 1; i <= n; ++i)
      if ((int)g[i].size() <= k - 1) q.emplace(g[i].size(), i);
    bool found_clique = false;
    while (!q.empty() && k * ll(k - 1) / 2 <= m) {
      int cur = q.top().second;
      q.pop();
      if (used[cur] == tst) continue;
      if ((int)g[cur].size() > k - 1) continue;
      if ((int)g[cur].size() == k - 1) {
        vector<int> neighbors = vector<int>(begin(g[cur]), end(g[cur]));
        neighbors.push_back(cur);
        random_shuffle(begin(neighbors), end(neighbors));
        bool good = true;
        for (int i = 0; i < k && good; ++i) {
          int ni = neighbors[i];
          for (int j = i + 1; j < k; ++j) {
            int nj = neighbors[j];
            if (g[ni].find(nj) == g[ni].end()) {
              good = false;
              break;
            }
          }
        }
        if (good) {
          found_clique = true;
          cout << "2\n";
          for (int x : neighbors) cout << x << ' ';
          cout << "\n";
          break;
        }
      }
      used[cur] = tst;
      for (int to : g[cur]) {
        g[to].erase(cur);
        q.emplace(g[to].size(), to);
        --m;
      }
      g[cur].clear();
    }
    if (found_clique) continue;
    vector<int> ans;
    for (int i = 1; i <= n; ++i) {
      if (used[i] != tst) ans.push_back(i);
    }
    if (ans.empty() || k * ll(k - 1) / 2 > m)
      cout << "-1\n";
    else {
      cout << "1 " << ans.size() << "\n";
      for (int x : ans) cout << x << ' ';
      cout << "\n";
    }
  }
}