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;
int n, m, k;
struct Edge {
  int to, next;
} e[100010 * 2];
int c1, head[100010], du[100010];
inline void add_edge(int u, int v) {
  e[c1].to = v;
  e[c1].next = head[u];
  head[u] = c1++;
}
vector<int> E[100010];
bool check(vector<int> &a) {
  for (auto u : a)
    for (auto v : a) {
      if (u == v) break;
      if (!binary_search(E[u].begin(), E[u].end(), v)) return 0;
    }
  return 1;
}
int vis[100010];
void init() {
  memset(head, -1, sizeof head);
  for (int i = 1; i <= n; ++i) head[i] = -1, E[i].clear(), du[i] = vis[i] = 0;
  c1 = 0;
}
void work() {
  cin >> n >> m >> k;
  init();
  for (int i = 1; i <= m; ++i) {
    int x, y;
    cin >> x >> y;
    add_edge(x, y);
    add_edge(y, x);
    ++du[x];
    ++du[y];
    E[x].push_back(y);
    E[y].push_back(x);
  }
  queue<int> Q;
  for (int i = 1; i <= n; ++i) sort(E[i].begin(), E[i].end());
  for (int i = 1; i <= n; ++i)
    if (du[i] < k) Q.push(i);
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    vis[u] = 1;
    if (du[u] == k - 1 && 1ll * k * (k - 1) / 2 <= m) {
      vector<int> clique = {u};
      for (int i = head[u]; ~i; i = e[i].next) {
        int v = e[i].to;
        if (vis[v]) continue;
        clique.push_back(v);
      }
      if (check(clique)) {
        cout << 2 << "\n";
        for (auto u : clique) cout << u << " ";
        cout << "\n";
        return;
      }
    }
    for (int i = head[u]; ~i; i = e[i].next) {
      int v = e[i].to;
      if (--du[v] == k - 1) Q.push(v);
    }
  }
  int s = 0;
  for (int i = 1; i <= n; ++i) s += du[i] >= k;
  if (s) {
    cout << 1 << " " << s << endl;
    for (int i = 1; i <= n; ++i)
      if (du[i] >= k) cout << i << " ";
    cout << "\n";
    return;
  }
  cout << "-1\n";
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int T;
  cin >> T;
  while (T--) work();
  return 0;
}