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("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
int n, m, k;
vector<set<int>> adj;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
bool check(vector<int>& sub) {
  auto ord = sub;
  shuffle(ord.begin(), ord.end(), rng);
  for (int& u : ord) {
    if (adj[u].size() > 50 * k) {
      for (int& v : ord)
        if (u != v && adj[u].find(v) == adj[u].end()) return false;
    } else {
      int ptr = 0;
      for (auto& v : adj[u]) {
        while (ptr < sub.size() && (v == sub[ptr] || u == sub[ptr])) ptr++;
        if (ptr == sub.size()) break;
      }
      if (ptr != sub.size()) return false;
    }
  }
  return true;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    adj.clear();
    scanf("%d%d%d", &n, &m, &k);
    adj.resize(n + 1);
    for (int a, b, i = 0; i < m; i++) {
      scanf("%d%d", &a, &b);
      adj[a].insert(b);
      adj[b].insert(a);
    }
    if (1LL * k * (k - 1) / 2 > m) {
      printf("-1\n");
      continue;
    }
    set<pair<int, int>> ver;
    for (int i = 1; i <= n; i++) ver.insert({adj[i].size(), i});
    bool ok = false;
    while (!ver.empty() && ver.begin()->first <= k - 1) {
      int u = ver.begin()->second, sz = ver.begin()->first;
      ver.erase(ver.begin());
      if (sz == k - 1) {
        vector<int> clique{adj[u].begin(), adj[u].end()};
        clique.insert(lower_bound(clique.begin(), clique.end(), u), u);
        if (check(clique)) {
          printf("2\n");
          for (int& i : clique) printf("%d ", i);
          printf("\n");
          ok = true;
          break;
        }
      }
      for (auto& v : adj[u]) {
        ver.erase({adj[v].size(), v});
        adj[v].erase(u);
        ver.insert({adj[v].size(), v});
      }
      adj[u].clear();
    }
    if (!ok && !ver.empty()) {
      printf("1 %d\n", ver.size());
      for (auto& p : ver) printf("%d ", p.second);
      printf("\n");
    } else if (!ok)
      printf("-1\n");
  }
}