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 <typename T>
void _do(T x) {
  cerr << x << "\n";
}
template <typename T, typename... U>
void _do(T x, U... y) {
  cerr << x << ", ";
  _do(y...);
}
int n, m, k, deg[100005], dead[100005];
unordered_set<int> E[100005];
bool check_clique(vector<long long> &v) {
  for (auto i : v) {
    for (auto j : v) {
      if (i == j) continue;
      if (!E[i].count(j)) return false;
    }
  }
  return true;
}
void solve() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) E[i].clear(), deg[i] = dead[i] = 0;
  for (int i = 1, a, b; i <= m; i++)
    cin >> a >> b, E[a].insert(b), E[b].insert(a), deg[a]++, deg[b]++;
  queue<int> del;
  for (int i = 1; i <= n; i++) {
    if (deg[i] < k - 1) del.push(i), dead[i] = 1;
  }
  while (!del.empty()) {
    int nd = del.front();
    del.pop();
    for (auto &i : E[nd]) {
      if (dead[i]) continue;
      E[i].erase(nd);
      deg[i]--;
      if (deg[i] < k - 1 && !dead[i]) del.push(i), dead[i] = 1;
    }
    E[nd].clear();
    deg[nd] = 0;
  }
  for (int i = 1; i <= n; i++)
    if (deg[i] == k - 1) del.push(i);
  while (!del.empty()) {
    int nd = del.front();
    del.pop();
    if (dead[nd]) continue;
    vector<long long> cur(1, nd);
    for (auto &i : E[nd])
      if (!dead[i]) cur.emplace_back(i);
    if (cur.size() == k && check_clique(cur)) {
      cout << 2 << '\n';
      for (auto i : cur) cout << i << " ";
      cout << '\n';
      return;
    }
    for (auto &i : E[nd]) {
      if (dead[i]) continue;
      E[i].erase(nd);
      deg[i]--;
      if (deg[i] < k && !dead[i]) del.push(i);
    }
    dead[nd] = 1;
    deg[nd] = 0;
    E[nd].clear();
  }
  vector<long long> ans;
  for (int i = 1; i <= n; i++)
    if (!dead[i]) ans.emplace_back(i);
  if (ans.size() >= k) {
    cout << 1 << " " << ans.size() << '\n';
    for (auto i : ans) cout << i << " ";
    cout << '\n';
  } else
    cout << -1 << '\n';
}
signed main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}