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;
constexpr int mod = 1e9 + 7;
constexpr int KINF = 0x3f3f3f3f;
constexpr double eps = 1e-7;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int _;
  cin >> _;
  while (_--) {
    int n, m, k;
    cin >> n >> m >> k;
    vector<vector<int> > g(n);
    vector<int> deg(n);
    vector<int> mark(n, 1);
    for (int i = (int)0; i < (int)m; ++i) {
      int u, v;
      cin >> u >> v;
      --u, --v;
      g[u].emplace_back(v);
      g[v].emplace_back(u);
      if (++deg[u] >= k) mark[u] = 0;
      if (++deg[v] >= k) mark[v] = 0;
    }
    queue<int> q;
    for (int i = (int)0; i < (int)n; ++i) {
      sort(g[i].begin(), g[i].end());
      if (deg[i] < k) q.push(i);
    }
    vector<int> clique;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      mark[u] = 2;
      if (clique.empty()) {
        clique = {u};
        bool ok = true;
        for (int& v : g[u])
          if (mark[v] != 2 && deg[v] >= k - 1) clique.emplace_back(v);
        for (int& x : clique)
          for (int& y : clique) {
            if (x == y) break;
            if (!binary_search(g[x].begin(), g[x].end(), y)) ok = false;
          }
        if (!ok || clique.size() != k) clique.clear();
      }
      for (int& v : g[u])
        if (--deg[v] < k && !mark[v]) {
          mark[v] = 1;
          q.push(v);
        }
    }
    int cnt = count(mark.begin(), mark.end(), 0);
    if (cnt) {
      cout << 1 << " " << cnt << endl;
      for (int i = (int)0; i < (int)n; ++i)
        if (!mark[i]) cout << i + 1 << " ";
      cout << endl;
    } else if (!clique.empty()) {
      cout << 2 << endl;
      for (int& i : clique) cout << i + 1 << " ";
      cout << endl;
    } else
      cout << -1 << endl;
  }
  return 0;
}