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 int N = 2e5 + 5;
vector<int> g[N];
vector<pair<int, int>> e[N];
int vis[N], deg[N], ord[N], dro[N], ans[N], orn;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, m, k;
    cin >> n >> m >> k;
    orn = 0;
    for (int i = 1; i <= n; i++) {
      g[i].clear();
      e[i].clear();
      vis[i] = 0;
      deg[i] = 0;
      ord[i] = 0;
      dro[i] = 0;
      ans[i] = 0;
    }
    for (int i = 1; i <= m; i++) {
      int u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
      deg[u]++;
      deg[v]++;
    }
    set<pair<int, int>> st;
    for (int i = 1; i <= n; i++) st.insert({deg[i], i});
    int cd = 0;
    while (!st.empty()) {
      int now = st.begin()->second;
      if (deg[now] >= k) break;
      ord[now] = ++orn;
      dro[orn] = now;
      st.erase(st.begin());
      vis[now] = 1;
      for (int i : g[now])
        if (vis[i] == 0) {
          st.erase({deg[i], i});
          st.insert({deg[i] -= 1, i});
        }
    }
    if (st.size() > k) {
      cout << 1 << ' ' << st.size() << endl;
      for (auto i : st) cout << i.second << ' ';
      cout << endl;
      continue;
    }
    for (int i = 1; i <= n; i++) vis[i] = 0;
    for (int i = 1; i <= orn; i++) {
      int now = dro[i];
      vector<int> nei;
      for (int j : g[now])
        if (ord[j] > ord[now]) nei.push_back(j);
      for (int j : nei) vis[j] = 1;
      for (auto j : e[now])
        if (vis[j.first] == 0) ans[j.second] = 0;
      for (int j : nei) vis[j] = 0;
      if (nei.size() != k - 1) continue;
      ans[now] = 1;
      for (int j = 0; j < nei.size(); j++)
        for (int k = 0; k < nei.size(); k++)
          if (ord[nei[j]] < ord[nei[k]]) e[nei[j]].push_back({nei[k], now});
    }
    bool ok2 = 0;
    for (int i = 1; i <= n; i++)
      if (ans[i]) {
        cout << 2 << endl;
        cout << i;
        for (int j : g[i])
          if (ord[j] > ord[i]) cout << ' ' << j;
        cout << endl;
        ok2 = 1;
        break;
      }
    if (ok2) continue;
    cout << -1 << endl;
  }
  return 0;
}