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;
void debug() { cerr << "\n"; }
template <typename H, typename... T>
void debug(H h, T... t) {
  cerr << h;
  if (sizeof...(t)) cerr << ", ";
  debug(t...);
}
const int N = 1e5 + 5, INF = 1e9 + 5, mod = 1e9 + 7;
unordered_set<int> E[N];
int d[N];
bool check(int v) {
  vector<int> V;
  for (auto &i : E[v]) V.push_back(i);
  for (int i = 0; i < (int)(V).size(); i++) {
    for (int j = i + 1; j < (int)(V).size(); j++) {
      if (E[V[i]].find(V[j]) == E[V[i]].end()) return 0;
    }
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n, m, k;
    cin >> n >> m >> k;
    for (int i = 0; i < m; i++) {
      int u, v;
      cin >> u >> v;
      d[u]++;
      d[v]++;
      E[v].insert(u);
      E[u].insert(v);
    }
    vector<int> zle;
    for (int i = 1; i <= n; i++) {
      if (d[i] < k) zle.push_back(i);
    }
    for (int i = 0; i < zle.size(); i++) {
      int v = zle[i];
      for (int u : E[v]) {
        if (d[u]-- == k) zle.push_back(u);
      }
    }
    if (zle.size() < n) {
      cout << "1 " << n - zle.size() << "\n";
      for (int i = 1; i <= n; i++) {
        if (d[i] >= k) cout << i << " ";
      }
      for (int i = 1; i <= n; i++) {
        d[i] = 0;
        E[i].clear();
      }
      cout << "\n";
      continue;
    }
    zle.clear();
    vector<int> V;
    for (int i = 1; i <= n; i++) {
      if (E[i].size() < k - 1) {
        zle.push_back(i);
      }
      if (E[i].size() == k - 1) V.push_back(i);
    }
    while (zle.size() || (int)(V).size()) {
      while (zle.size()) {
        int v = zle.back();
        zle.pop_back();
        for (int u : E[v]) {
          E[u].erase(E[u].find(v));
          if (E[u].size() == k - 1) {
            V.push_back(u);
          }
          if (E[u].size() == k - 2) {
            zle.push_back(u);
          }
        }
        E[v].clear();
      }
      if (V.size()) {
        int v = V.back();
        V.pop_back();
        if (E[v].size() < k - 1) continue;
        if (check(v)) {
          cout << 2 << "\n";
          cout << v << " ";
          for (int i : E[v]) cout << i << " ";
          cout << "\n";
          V.push_back(0);
          break;
        } else
          zle.push_back(v);
      }
    }
    if (!(int)(V).size()) cout << "-1\n";
    for (int i = 1; i <= n; i++) {
      d[i] = 0;
      E[i].clear();
    }
  }
}