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;
const long long Mod = 1000000007LL;
const int N = 1e5 + 10;
const long long Inf = 2242545357980376863LL;
const long long Log = 30;
int adj[N];
int deg[N], mk[N];
vector<int> V[N], G[N], R[N];
int la = -1;
int val[N];
int n, m, k;
set<pair<int, int> > st;
int Main() {
  st.clear();
  while (la >= 0) {
    V[la].clear();
    la--;
  }
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) deg[i] = 0, G[i].clear(), R[i].clear();
  int u, v;
  for (int i = 1; i <= m; i++) {
    cin >> u >> v;
    deg[u]++;
    deg[v]++;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    mk[i] = 1;
    st.insert({deg[i], i});
  }
  while (!st.empty()) {
    pair<int, int> fr = *st.begin();
    if (fr.first >= k) {
      cout << "1 " << st.size() << '\n';
      for (auto X : st) cout << X.second << ' ';
      cout << '\n';
      return 0;
    }
    if (fr.first == k - 1) {
      la++;
      V[la].push_back(fr.second);
      for (auto nei : G[fr.second]) {
        if (mk[nei]) V[la].push_back(nei);
      }
    }
    st.erase(st.begin());
    mk[fr.second] = 0;
    for (auto nei : G[fr.second]) {
      if (mk[nei]) {
        st.erase({deg[nei], nei});
        deg[nei]--;
        st.insert({deg[nei], nei});
      }
    }
  }
  if ((k * (k - 1)) > m + m) {
    cout << "-1\n";
    return 0;
  }
  for (int i = 0; i <= la; i++) {
    val[i] = 1;
    for (auto nei : V[i]) R[nei].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    for (auto nei : G[i]) {
      adj[nei] = 1;
    }
    adj[i] = 1;
    for (int al : R[i]) {
      if (!val[al]) continue;
      for (auto el : V[al]) {
        if (adj[el] == 0) {
          val[al] = 0;
          break;
        }
      }
    }
    adj[i] = 0;
    for (auto nei : G[i]) {
      adj[nei] = 0;
    }
  }
  for (int i = 0; i <= la; i++) {
    if (val[i]) {
      cout << "2\n";
      for (auto x : V[i]) cout << x << ' ';
      cout << '\n';
      return 0;
    }
  }
  cout << "-1\n";
  return 0;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long T;
  cin >> T;
  while (T--) {
    Main();
  }
  return 0;
}