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;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c *x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  ~debug() { cerr << endl; }
  template <class c>
  typename enable_if<sizeof dud<c>(0) != 1, debug &>::type operator<<(c i) {
    cerr << boolalpha << i;
    return *this;
  }
  template <class c>
  typename enable_if<sizeof dud<c>(0) == 1, debug &>::type operator<<(c i) {
    return *this << range(begin(i), end(i));
  }
  template <class c, class b>
  debug &operator<<(pair<b, c> d) {
    return *this << "(" << d.first << ", " << d.second << ")";
  }
  template <class c>
  debug &operator<<(rge<c> d) {
    *this << "[";
    for (auto it = d.b; it != d.e; ++it) *this << ", " + 2 * (it == d.b) << *it;
    return *this << "]";
  }
};
const long double eps = 1e-9;
const int T = 1e5;
const int M = 1e9 + 7;
const long long inf = 1e18;
auto rnd = bind(uniform_int_distribution<int>(1, 5), mt19937(time(0)));
const int N = 2e5 + 1;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  ;
  int t;
  cin >> t;
  while (t--) {
    int n, m, k;
    cin >> n >> m >> k;
    set<pair<int, int>> st;
    vector<vector<int>> g(n, vector<int>());
    vector<int> deg(n, 0);
    vector<bool> used(n, 0);
    for (int i = 0; i < m; i++) {
      int a, b;
      cin >> a >> b;
      --a;
      --b;
      g[a].push_back(b);
      g[b].push_back(a);
    }
    for (int i = 0; i < n; i++) {
      sort(g[i].begin(), g[i].end());
      deg[i] = (int)g[i].size();
      st.insert({deg[i], i});
      used[i] = 0;
    }
    bool answered = 0;
    while (!st.empty() && !answered) {
      pair<int, int> cur = *st.begin();
      int cl = cur.first, v = cur.second;
      if (cl >= k) break;
      st.erase(st.begin());
      used[v] = 1;
      for (auto &z : g[v]) {
        if (used[z]) continue;
        st.erase({deg[z], z});
        deg[z]--;
        st.insert({deg[z], z});
      }
    }
    if ((int)st.size()) {
      cout << 1 << ' ' << (int)st.size() << '\n';
      for (auto &z : st) cout << z.second + 1 << ' ';
      cout << '\n';
      continue;
    }
    if (1ll * k * (k - 1) / 2 > m) {
      cout << -1 << '\n';
      continue;
    }
    for (int i = 0; i < n; i++) {
      deg[i] = (int)g[i].size();
      st.insert({deg[i], i});
      used[i] = 0;
    }
    while ((int)st.size() && !answered) {
      pair<int, int> cur = *st.begin();
      st.erase(st.begin());
      int cl = cur.first, v = cur.second;
      if (cl == (k - 1)) {
        bool ok = 1;
        vector<int> w;
        w.push_back(v);
        for (auto &z : g[v])
          if (!used[z]) w.push_back(z);
        for (int i = 0; ok && i < (int)w.size(); i++) {
          for (int j = i + 1; ok && j < (int)w.size(); j++) {
            int first = w[i], second = w[j];
            ok &= binary_search(g[first].begin(), g[first].end(), second);
          }
        }
        if (ok) {
          cout << 2 << '\n';
          for (auto &z : w) cout << z + 1 << ' ';
          cout << '\n';
          answered = 1;
          break;
        }
      }
      used[v] = 1;
      for (auto &z : g[v]) {
        if (used[z]) continue;
        st.erase({deg[z], z});
        deg[z]--;
        st.insert({deg[z], z});
      }
    }
    if (!answered) cout << -1 << '\n';
  }
  return 0;
}