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 ll = long long;
using uint = unsigned int;
template <class T>
using V = vector<T>;
template <class T>
using VV = vector<vector<T>>;
template <class T, class U>
void chmax(T& x, U y) {
  if (x < y) x = y;
}
template <class T, class U>
void chmin(T& x, U y) {
  if (y < x) x = y;
}
template <class T>
void mkuni(V<T>& v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
template <class S, class T>
ostream& operator<<(ostream& o, const pair<S, T>& p) {
  return o << "(" << p.first << "," << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& vc) {
  o << "{";
  for (const T& v : vc) o << v << ",";
  o << "}";
  return o;
}
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n - 1); }
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);
  int T;
  cin >> T;
  while (T--) {
    int N, M, K;
    cin >> N >> M >> K;
    using P = pair<int, int>;
    set<P> st;
    VV<int> G(N);
    V<int> deg(N);
    for (int i = 0; i < int(M); i++) {
      int x, y;
      cin >> x >> y;
      x--, y--;
      G[x].emplace_back(y);
      G[y].emplace_back(x);
      deg[x]++, deg[y]++;
    }
    for (int i = 0; i < int(N); i++) {
      sort(G[i].begin(), G[i].end());
      st.insert(P(deg[i], i));
    }
    if (1LL * K * (K - 1) / 2 > M) {
      cout << -1 << endl;
      continue;
    }
    bool ok = false;
    V<int> buf(K);
    while (!st.empty()) {
      P p = *st.begin();
      st.erase(p);
      int v = p.second;
      if (deg[v] >= K) {
        cout << 1 << " " << int(st.size()) + 1 << endl;
        cout << v + 1 << " ";
        for (P p : st) cout << p.second + 1 << " ";
        cout << endl;
        ok = true;
        break;
      }
      if (deg[v] < K - 1) {
        for (int u : G[v])
          if (deg[u] != -1) {
            st.erase(P(deg[u], u));
            deg[u]--;
            st.insert(P(deg[u], u));
          }
        deg[v] = -1;
        continue;
      }
      {
        buf[0] = v;
        int I = 1;
        for (int u : G[v])
          if (deg[u] != -1) {
            buf[I++] = u;
          }
        assert(I == K);
        auto clique = [&]() {
          for (int i = 0; i < int(K); i++)
            for (int j = 0; j < int(i); j++) {
              if (!binary_search(G[buf[i]].begin(), G[buf[i]].end(), buf[j]))
                return false;
            }
          return true;
        };
        if (clique()) {
          cout << 2 << endl;
          for (int v : buf) cout << v + 1 << " ";
          cout << endl;
          ok = true;
          break;
        }
        for (int u : G[v])
          if (deg[u] != -1) {
            st.erase(P(deg[u], u));
            deg[u]--;
            st.insert(P(deg[u], u));
          }
        deg[v] = -1;
        continue;
      }
    }
    if (!ok) {
      cout << -1 << endl;
    }
  }
}