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 = 1e5 + 10;
int vis[N];
int cnt[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, m, k;
    memset(vis, 0, sizeof(vis));
    cin >> n >> m >> k;
    vector<vector<int> > ed(n + 1);
    for (int i = 0; i < m; i++) {
      int x, y;
      cin >> x >> y;
      ed[x].push_back(y);
      ed[y].push_back(x);
    }
    queue<int> q;
    for (int i = 1; i <= n; i++) {
      sort(ed[i].begin(), ed[i].end());
      cnt[i] = ed[i].size();
      if (cnt[i] < k) vis[i] = 1, q.push(i);
    }
    vector<int> ans;
    while (!q.empty()) {
      int x = q.front();
      q.pop();
      vis[x] = 2;
      if (cnt[x] == k - 1 && ans.size() == 0 && k * (k - 1) / 2 <= m) {
        ans.push_back(x);
        for (auto y : ed[x])
          if (vis[y] < 2) ans.push_back(y);
        bool flag = 1;
        for (auto i : ans)
          for (auto j : ans)
            if (i != j && !binary_search(ed[i].begin(), ed[i].end(), j))
              flag = 0;
        if (!flag) ans.clear();
      }
      for (int i = 0; i < ed[x].size(); i++) {
        int y = ed[x][i];
        if (--cnt[y] < k && !vis[y]) vis[y] = 1, q.push(y);
      }
    }
    int cn = 0;
    for (int i = 1; i <= n; i++)
      if (!vis[i]) cn++;
    if (cn > 0) {
      cout << 1 << " " << cn << endl;
      bool f = 0;
      for (int i = 1; i <= n; i++) {
        if (!vis[i]) {
          if (f) cout << " ";
          f = 1;
          cout << i;
        }
      }
      cout << endl;
    } else if (ans.size()) {
      cout << 2 << endl;
      for (int i = 0; i < ans.size(); i++) {
        if (i != 0) cout << " ";
        cout << ans[i];
      }
      cout << endl;
    } else
      cout << -1 << endl;
  }
}