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;
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};
const int MAX = 1e5 + 5;
int N, M, K;
vector<int> G[MAX];
int in[MAX];
bool vis[MAX], del[MAX];
bool chk(int x) {
  vector<int> v;
  v.emplace_back(x);
  for (auto& y : G[x])
    if (!del[y]) v.emplace_back(y);
  for (int i = 0; i < (int)(v.size()); i++) {
    for (int j = i + 1; j < (int)(v.size()); j++) {
      int a = v[i], b = v[j];
      if (!binary_search((G[a]).begin(), (G[a]).end(), b)) return 0;
    }
  }
  return 1;
}
void solve() {
  cin >> N >> M >> K;
  for (int i = 1; i <= N; i++) {
    G[i].clear();
    vis[i] = in[i] = del[i] = 0;
  }
  for (int i = 1; i <= M; i++) {
    int u, v;
    cin >> u >> v;
    G[u].emplace_back(v);
    G[v].emplace_back(u);
    in[u]++, in[v]++;
  }
  for (int i = 1; i <= N; i++) sort((G[i]).begin(), (G[i]).end());
  queue<int> q;
  for (int i = 1; i <= N; i++) {
    if (in[i] <= K - 1) {
      q.push(i);
      vis[i] = 1;
    }
  }
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    if (1LL * K * (K - 1) <= 2 * M && in[x] == K - 1) {
      if (chk(x)) {
        cout << 2 << '\n';
        cout << x << ' ';
        for (auto& y : G[x])
          if (!del[y]) cout << y << ' ';
        cout << '\n';
        return;
      }
    }
    del[x] = 1;
    for (auto& y : G[x]) {
      in[y]--;
      if (in[y] <= K - 1 && !vis[y]) q.push(y), vis[y] = 1;
    }
  }
  vector<int> v;
  for (int i = 1; i <= N; i++)
    if (!del[i]) v.push_back(i);
  if ((int)(v.size())) {
    cout << 1 << ' ' << (int)(v.size()) << '\n';
    for (auto& x : v) cout << x << ' ';
    cout << '\n';
    return;
  }
  cout << -1 << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T;
  cin >> T;
  while (T--) solve();
  return 0;
}