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;
string to_string(const char* s) { return (string)s; }
string to_string(string s) { return s; }
template <class T>
string to_string(T v) {
  string res = "";
  for (const auto& x : v) res += to_string(x) + " ";
  if (res.size()) res.pop_back();
  return res;
}
template <class A>
void pr(A x) {
  cout << to_string(x);
}
template <class H, class... T>
void pr(const H& h, const T&... t) {
  pr(h);
  pr(t...);
}
void ps() { pr("\n"); }
template <class H, class... T>
void ps(const H& h, const T&... t) {
  pr(h);
  if (sizeof...(t)) pr(" ");
  ps(t...);
}
class number_iterator : public iterator<forward_iterator_tag, int> {
 public:
  int v;
  int inc;
  number_iterator(int v, int inc = 1) : v(v), inc(inc) {}
  operator int&() { return v; }
  int operator*() { return v; }
  number_iterator operator++() { return v += inc; }
};
class range : pair<int, int> {
  int inc;
  void init(int inc = 1) {
    this->second = max(this->first, this->second);
    int div = (this->second - this->first) / inc;
    if ((this->second - this->first) % inc != 0) div += 1;
    this->second = this->first + div * inc;
  }
  void init_dec(int dec = 1) {
    this->first = max(this->first, this->second);
    int div = (this->first - this->second) / dec;
    if ((this->first - this->second) % dec != 0) div += 1;
    this->second = this->first - div * dec;
  }

 public:
  range(int first, int second, int inc = 1)
      : pair<int, int>(first, second), inc(inc) {
    if (inc < 0)
      init_dec(-inc);
    else
      init(inc);
  }
  range(int n) : range(0, n, 1) {}
  number_iterator begin() { return number_iterator(first, inc); }
  number_iterator end() { return number_iterator(second, inc); }
  operator vector<int>() { return vector<int>(this->begin(), this->end()); }
};
class graph {
 protected:
  int n;

 public:
  vector<unordered_set<int>> g;
  graph(int n) : n(n) { g.resize(n); }
  void add(int from, int to) {
    g[from].insert(to);
    g[to].insert(from);
  }
};
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  long long _min = 1ll * k * (k - 1);
  graph g(n);
  for (int i : range(m)) {
    int u, w;
    cin >> u >> w;
    g.add(--u, --w);
  }
  if (2 * m < _min) return ps(-1);
  set<pair<int, int>> q;
  for (int i : range(n)) q.insert({g.g[i].size(), i});
  while (q.size()) {
    pair<int, int> cur = *q.begin();
    if (cur.first < k - 1) {
      for (int i : g.g[cur.second]) {
        q.erase({g.g[i].size(), i});
        g.g[i].erase(cur.second);
        q.insert({g.g[i].size(), i});
      }
    } else if (cur.first > k - 1) {
      ps(1, q.size());
      for (auto& p : q) pr(p.second + 1, " ");
      return ps();
    } else {
      bool flag = true;
      auto& l = g.g[cur.second];
      int skip = 0;
      for (int x : l) {
        skip++;
        if (flag) {
          int skip_l = skip;
          for (int y : g.g[cur.second]) {
            if (skip_l) {
              skip_l--;
              continue;
            }
            if (!g.g[x].count(y)) {
              flag = false;
              break;
            }
          }
        }
      }
      if (flag) {
        ps(2);
        pr(cur.second + 1, " ");
        for (auto& v : g.g[cur.second]) pr(v + 1, " ");
        return ps();
      }
      for (int i : g.g[cur.second]) {
        q.erase({g.g[i].size(), i});
        g.g[i].erase(cur.second);
        q.insert({g.g[i].size(), i});
      }
    }
    q.erase(cur);
  }
  ps("-1");
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) solve();
}