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 T1, class T2>
ostream& operator<<(ostream& out, const pair<T1, T2> p) {
  out << '(' << p.first << ',' << p.second << ')';
  return out;
}
template <class T1, class T2>
istream& operator>>(istream& in, pair<T1, T2>& p) {
  in >> p.first >> p.second;
  return in;
}
template <class T>
istream& operator>>(istream& in, vector<T>& v) {
  for (T& x : v) in >> x;
  return in;
}
template <class T>
ostream& operator<<(ostream& out, const vector<vector<T>>& v) {
  for (const vector<T>& x : v) out << x << '\n';
  return out;
}
template <class T>
ostream& operator<<(ostream& out, const vector<T>& v) {
  for (const T& x : v) out << x << ' ';
  return out;
}
long long gcd(long long a, long long b) {
  if (b > a) swap(a, b);
  return (b ? gcd(b, a % b) : a);
}
using ll = long long;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using tiii = pair<pair<int, int>, int>;
using vi = vector<int>;
using vl = vector<long long>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<long long>>;
const int h = 1000000007;
class Dsu {
 private:
  vector<int> par, size;
  int distinct_sets_count;
  int root(const int x) {
    if (par[x] == x) return x;
    return par[x] = root(par[x]);
  }

 public:
  Dsu(int n) {
    distinct_sets_count = n;
    par.resize(n);
    for (int i = 0; i < n; i++) par[i] = i;
    size.resize(n, 1);
  }
  void connect(int a, int b) {
    a = root(a), b = root(b);
    if (a == b) return;
    if (size[a] < size[b]) swap(a, b);
    size[a] += size[b];
    par[b] = a;
    distinct_sets_count--;
  }
  int same_component(const int x, const int y) { return root(x) == root(y); }
  int size_of_component(const int x) { return size[root(x)]; }
  int count() const { return distinct_sets_count; }
};
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cerr.setstate(ios::failbit);
  int t;
  cin >> t;
  while (t--) {
    int n, m, k;
    cin >> n >> m >> k;
    vector<set<int>> M(n);
    for (int i = (0); i < (m); i++) {
      int x, y;
      cin >> x >> y;
      x--, y--;
      M[x].insert(y);
      M[y].insert(x);
    }
    auto M2 = M;
    bool found = false;
    int comp = -1;
    for (int _ = (0); _ < (2); _++) {
      M = M2;
      stack<int> st;
      for (int i = (0); i < (n); i++)
        if (M[i].size() < k) st.push(i);
      while (!st.empty()) {
        int x = st.top();
        st.pop();
        for (int y : M[x]) {
          M[y].erase(x);
          if (M[y].size() < k) st.push(y);
        }
        M[x].clear();
      }
      Dsu dsu(n);
      for (int i = (0); i < (n); i++) {
        for (int y : M[i]) dsu.connect(i, y);
      }
      vi ans;
      for (int i = (0); i < (n); i++) {
        if (_ == 0 and dsu.size_of_component(i) > 1) {
          found = true;
          comp = i;
        } else if (_ == 1 and M[i].size() == k) {
          st.push(i);
        }
      }
      if (_ == 1) {
        while (!st.empty()) {
          int i = st.top();
          st.pop();
          if (M[i].size() == k) {
            vi v{i};
            bool ok = true;
            for (int y : M[i]) {
              v.push_back(y);
            }
            for (int i = (0); i < (v.size()); i++) {
              for (int j = (i + 1); j < (v.size()); j++) {
                if (!M[v[i]].count(v[j])) {
                  ok = false;
                  break;
                }
              }
              if (!ok) break;
            }
            if (ok) {
              found = true;
              ans = v;
              break;
            }
          }
          for (int y : M[i]) {
            M[y].erase(i);
            if (M[y].size() == k) st.push(y);
          }
          M[i].clear();
        }
      }
      if (found) {
        if (_ == 0)
          for (int i = (0); i < (n); i++) {
            if (dsu.same_component(comp, i)) ans.push_back(i);
          }
        cout << _ + 1;
        if (_ + 1 == 1) cout << ' ' << ans.size();
        cout << '\n';
        for (int i = (0); i < (ans.size()); i++) cout << ans[i] + 1 << ' ';
        cout << '\n';
        break;
      }
      k--;
    }
    if (!found) cout << "-1\n";
  }
}