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;
vector<int> edges[200005];
int deg[200005];
bool vis[200005], excl[200005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m, i, j, u, v, w, k, t;
  cin >> t;
  while (t--) {
    cin >> n >> m >> k;
    for (i = 1; i <= n; i++) {
      deg[i] = 0;
      edges[i].clear();
      vis[i] = false;
      excl[i] = false;
    }
    for (i = 1; i <= m; i++) {
      cin >> u >> v;
      edges[u].push_back(v);
      edges[v].push_back(u);
      deg[u]++;
      deg[v]++;
    }
    for (i = 1; i <= n; i++) sort(edges[i].begin(), edges[i].end());
    priority_queue<pair<int, int>, vector<pair<int, int> >,
                   greater<pair<int, int> > >
        pq;
    for (i = 1; i <= n; i++) pq.push({deg[i], i});
    vector<int> ans;
    bool clique = false;
    while (!pq.empty()) {
      j = pq.top().first;
      u = pq.top().second;
      pq.pop();
      if (vis[u]) continue;
      vis[u] = true;
      if (j >= k) {
        ans.push_back(u);
        continue;
      } else if (j == (k - 1) && ((k - 1) * k / 2 <= m)) {
        int sz = edges[u].size();
        bool ase = true;
        for (i = 0; i < sz; i++) {
          v = edges[u][i];
          if (excl[v]) continue;
          for (j = i + 1; j < sz; j++) {
            w = edges[u][j];
            if (excl[w]) continue;
            ase = binary_search(edges[v].begin(), edges[v].end(), w);
            if (!ase) break;
          }
          if (!ase) break;
        }
        if (ase) {
          clique = true;
          break;
        }
      }
      excl[u] = true;
      for (int v : edges[u]) {
        deg[v]--;
        pq.push({deg[v], v});
      }
    }
    if (clique) {
      cout << 2 << endl << u << ' ';
      for (int v : edges[u]) {
        if (!excl[v]) cout << v << ' ';
      }
      cout << endl;
    } else if (ans.size() > k) {
      cout << 1 << ' ' << ans.size() << endl;
      for (int x : ans) cout << x << ' ';
      cout << endl;
    } else {
      cout << -1 << endl;
    }
  }
}