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;
using namespace std;
template <typename T, typename U>
std::istream& operator>>(std::istream& i, pair<T, U>& p) {
  i >> p.first >> p.second;
  return i;
}
template <typename T>
std::istream& operator>>(std::istream& i, vector<T>& t) {
  for (auto& v : t) {
    i >> v;
  }
  return i;
}
template <typename T, typename U>
std::ostream& operator<<(std::ostream& o, const pair<T, U>& p) {
  o << p.first << ' ' << p.second;
  return o;
}
template <typename T>
std::ostream& operator<<(std::ostream& o, const vector<T>& t) {
  if (t.empty()) o << '\n';
  for (size_t i = 0; i < t.size(); ++i) {
    o << t[i] << " \n"[i == t.size() - 1];
  }
  return o;
}
template <typename T>
using minheap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using maxheap = priority_queue<T, vector<T>, less<T>>;
template <typename T>
bool in(T a, T b, T c) {
  return a <= b && b < c;
}
unsigned int logceil(int x) {
  return (unsigned int)(8 * sizeof(int) - __builtin_clz(x));
}
namespace std {
template <typename T, typename U>
struct hash<pair<T, U>> {
  hash<T> t;
  hash<U> u;
  size_t operator()(const pair<T, U>& p) const {
    return t(p.x) ^ (u(p.y) << 7);
  }
};
}  // namespace std
template <typename T>
T gcd(T a, T b) {
  if (a < b) swap(a, b);
  return b ? gcd(b, a % b) : a;
}
template <typename T>
class vector2 : public vector<vector<T>> {
 public:
  vector2() {}
  vector2(size_t a, size_t b, T t = T())
      : vector<vector<T>>(a, vector<T>(b, t)) {}
};
template <typename T>
class vector3 : public vector<vector2<T>> {
 public:
  vector3() {}
  vector3(size_t a, size_t b, size_t c, T t = T())
      : vector<vector2<T>>(a, vector2<T>(b, c, t)) {}
};
template <typename T>
class vector4 : public vector<vector3<T>> {
 public:
  vector4() {}
  vector4(size_t a, size_t b, size_t c, size_t d, T t = T())
      : vector<vector3<T>>(a, vector3<T>(b, c, d, t)) {}
};
template <typename T>
class vector5 : public vector<vector4<T>> {
 public:
  vector5() {}
  vector5(size_t a, size_t b, size_t c, size_t d, size_t e, T t = T())
      : vector<vector4<T>>(a, vector4<T>(b, c, d, e, t)) {}
};
void fastIO() { ios::sync_with_stdio(false); }
vector<vector<int>> g;
vector<int> deg;
set<int> cur;
const int MAXK = 455;
int n, m, k;
bool check(int v) {
  if (k > MAXK) return false;
  vector<int> targ = {v};
  for (auto nb : g[v])
    if (cur.count(nb)) targ.push_back(nb);
  assert(targ.size() == k);
  for (auto cv : targ) {
    for (auto nb : targ) {
      if (nb == cv) continue;
      if (!binary_search(g[cv].begin(), g[cv].end(), nb)) return false;
    }
  }
  return true;
}
class BZadachaOPodmnozhestveGrafa {
 public:
  void solve(std::istream& in, std::ostream& out) {
    fastIO();
    int t;
    in >> t;
    for (int _ = (0); _ < (t); ++_) {
      in >> n >> m >> k;
      g.clear();
      g.resize(n);
      deg.clear();
      deg.resize(n);
      for (int i = (0); i < (m); ++i) {
        int u, v;
        in >> u >> v;
        --u;
        --v;
        g[u].push_back(v);
        g[v].push_back(u);
        deg[u]++;
        deg[v]++;
      }
      for (int i = (0); i < (n); ++i) sort(g[i].begin(), g[i].end());
      cur.clear();
      for (int i = (0); i < (n); ++i) cur.insert(i);
      set<pair<int, int>> tr;
      for (int i = (0); i < (n); ++i) tr.insert({deg[i], i});
      while (!tr.empty()) {
        auto fr = *tr.begin();
        if (fr.first >= k) {
          out << 1 << " " << cur.size() << endl;
          for (auto v : cur) out << v + 1 << " ";
          out << endl;
          break;
        }
        tr.erase(tr.begin());
        int v = fr.second;
        cur.erase(v);
        if (deg[v] == k - 1 && check(v)) {
          out << 2 << endl << v + 1 << " ";
          for (auto nb : g[v])
            if (cur.count(nb)) {
              out << nb + 1 << " ";
            }
          out << endl;
          break;
        }
        deg[v] = -1;
        for (auto nb : g[v]) {
          if (deg[nb] == -1) continue;
          tr.erase({deg[nb], nb});
          deg[nb]--;
          tr.insert({deg[nb], nb});
        }
      }
      if (tr.empty()) {
        out << -1 << endl;
      }
    }
  }
};
int main() {
  BZadachaOPodmnozhestveGrafa solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  cerr << 1.0 * clock() / CLOCKS_PER_SEC << endl;
  return 0;
}