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 + 6;
int T, deg[N], n, m, k;
vector<int> G[N];
bool vis[N];
bool solve1() {
  queue<int> q;
  for (int i = 1; i <= n; ++i) {
    vis[i] = 0;
    if (deg[i] < k) q.push(i);
  }
  while (q.size()) {
    int u = q.front();
    q.pop();
    vis[u] = 1;
    for (int v : G[u])
      if (deg[v]-- == k) q.push(v);
  }
  vector<int> vec;
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) vec.emplace_back(i);
  if (vec.size()) {
    cout << "1 " << vec.size() << '\n';
    for (int i : vec) cout << i << ' ';
    cout << '\n';
    return 1;
  }
  return 0;
}
bool solve2() {
  if (2 * m / (k - 1) < k) return 0;
  queue<int> q;
  for (int i = 1; i <= n; ++i) {
    deg[i] = G[i].size();
    vis[i] = 0;
    if (deg[i] < k - 1) q.push(i), vis[i] = 1;
  }
  while (q.size()) {
    int u = q.front();
    q.pop();
    for (int v : G[u])
      if (deg[v]-- == k - 1) q.push(v), vis[v] = 1;
  }
  for (int i = 1; i <= n; ++i)
    if (deg[i] == k - 1) q.push(i);
  while (q.size()) {
    int x = q.front();
    q.pop();
    vis[x] = 1;
    if (deg[x] == k - 1) {
      bool ok = 1;
      vector<int> vc;
      for (int i : G[x])
        if (!vis[i]) vc.emplace_back(i);
      for (int v = 0; v < (int)vc.size() && ok; ++v)
        for (int w = v + 1; w < (int)vc.size() && ok; ++w)
          if (!binary_search(G[vc[v]].begin(), G[vc[v]].end(), vc[w])) ok = 0;
      if (ok) {
        cout << "2\n" << x;
        for (int v : vc) cout << ' ' << v;
        cout << '\n';
        return 1;
      }
    }
    for (int v : G[x]) {
      if (deg[v]-- == k) q.push(v);
    }
  }
  return 0;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  for (cin >> T; T--;) {
    cin >> n >> m >> k;
    for (int i = 1; i <= m; ++i) {
      int u, v;
      cin >> u >> v;
      G[u].emplace_back(v), G[v].emplace_back(u);
      ++deg[u], ++deg[v];
    }
    for (int i = 1; i <= n; ++i) sort(G[i].begin(), G[i].end());
    if (!solve1() && !solve2()) cout << "-1\n";
    for (int i = 1; i <= n; ++i) deg[i] = 0, G[i].clear();
  }
  return 0;
}