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>
int n, m, k;
std::unordered_set<int> G[100123];
int deg[100123];
inline bool CheckClique(int x) {
  std::vector<int> V({x});
  for (int y : G[x]) V.emplace_back(y);
  for (int u : V)
    for (int v : V)
      if (u < v && (!G[u].count(v))) return false;
  return true;
}
inline void Solve() {
  for (int i = 1; i <= n; ++i) G[i].clear();
  memset(deg, 0, sizeof(deg));
  scanf("%d%d%d", &n, &m, &k);
  std::vector<std::pair<int, int>> input;
  for (int i = 1, u, v; i <= m; ++i) {
    scanf("%d%d", &u, &v);
    input.emplace_back(std::make_pair(u, v));
  }
  if ((1LL * k * (k - 1) / 2) > m) return puts("-1"), void();
  for (auto& p : input) {
    G[p.first].insert(p.second);
    G[p.second].insert(p.first);
    ++deg[p.first];
    ++deg[p.second];
  }
  int cnt = 0;
  std::queue<int> q;
  for (int i = 1; i <= n; ++i)
    if (deg[i] < k) {
      if (deg[i] == k - 1 && CheckClique(i)) {
        puts("2");
        for (int x : G[i]) printf("%d ", x);
        printf("%d\n", i);
        return;
      }
      q.push(i);
      ++cnt;
    }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int v : G[u]) {
      G[v].erase(u);
      if ((--deg[v]) == k - 1) {
        if (CheckClique(v)) {
          puts("2");
          for (int x : G[v]) printf("%d ", x);
          printf("%d\n", v);
          return;
        }
        q.push(v);
        ++cnt;
      }
    }
  }
  if (cnt < n) {
    std::vector<int> ans;
    for (int i = 1; i <= n; ++i)
      if (deg[i] >= k) ans.emplace_back(i);
    printf("1\n%d\n", static_cast<int>(ans.size()));
    for (int i : ans) printf("%d ", i);
    puts("");
  } else
    puts("-1");
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) Solve();
  return 0;
}