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 double eps = 1e-7;
const double PI = acos(-1);
const int MOD = 998244353;
const int INF = 0x3f3f3f3f;
const int N = 1e5 + 7;
int n, m, k;
int d[N], vis[N];
vector<int> e[N];
inline void solve() {
  queue<int> q;
  vector<int> clique;
  cin >> n >> m >> k;
  for (int i = 1, u, v; i <= m; ++i) {
    cin >> u >> v;
    e[u].emplace_back(v);
    e[v].emplace_back(u);
    ++d[u];
    ++d[v];
  }
  int r = n;
  for (int i = 1; i <= n; ++i) {
    if (d[i] < k) q.push(i);
    sort(e[i].begin(), e[i].end());
  }
  while (q.size()) {
    int u = q.front();
    q.pop();
    --r;
    vis[u] = 1;
    if (d[u] == k - 1 && k * (k - 1ll) / 2 <= m) {
      clique = vector<int>(1, u);
      for (const int &i : e[u])
        if (!vis[i]) clique.emplace_back(i);
      int flag = 1;
      for (int &i : clique) {
        for (int &j : clique) {
          if (i == j) break;
          auto it = lower_bound(e[i].begin(), e[i].end(), j);
          if (*it != j || it == e[i].end()) flag = 0;
        }
      }
      if (flag) {
        cout << 2 << endl;
        for (int &i : clique) cout << i << " ";
        cout << endl;
        return;
      }
    }
    for (const int &v : e[u]) {
      if (d[v]-- == k) q.push(v);
    }
  }
  if (r) {
    cout << 1 << " " << r << endl;
    for (int i = 1; i <= n; ++i)
      if (!vis[i]) {
        cout << i << " ";
      }
    cout << endl;
  } else {
    cout << -1 << endl;
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int T = 1;
  cin >> T;
  for (int t = 1; t <= T; ++t) {
    solve();
    for (int i = 1; i <= n; ++i) {
      e[i].clear();
      d[i] = vis[i] = 0;
    }
  }
  return 0;
}