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>
std::vector<int> vec[100010];
int r[100010];
int cnt[100010];
bool del[100010], inQueue[100010];
bool judge(int x) {
  std::vector<int> t;
  for (int i : vec[x])
    if (!del[i]) t.push_back(i);
  for (int i : t)
    if (i != x) {
      for (int j : t) {
        if (i == j) break;
        if (!std::binary_search(vec[i].begin(), vec[i].end(), j)) return false;
      }
    }
  return true;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, m, k, u, v;
    scanf("%d %d %d", &n, &m, &k);
    for (int i = 0; i < n; ++i) {
      vec[i].clear();
      inQueue[i] = del[i] = false;
    }
    for (int i = 0; i < m; ++i) {
      scanf("%d %d", &u, &v);
      --u;
      --v;
      vec[u].push_back(v);
      vec[v].push_back(u);
    }
    std::priority_queue<std::pair<int, int>, std::vector<std::pair<int, int>>,
                        std::greater<std::pair<int, int>>>
        que;
    for (int i = 0; i < n; ++i) {
      vec[i].push_back(i);
      std::sort(vec[i].begin(), vec[i].end());
      cnt[i] = vec[i].size();
      if (cnt[i] <= k) que.push(std::pair<int, int>(cnt[i], i));
    }
    bool flag = false;
    while (!que.empty()) {
      std::pair<int, int> x = que.top();
      que.pop();
      if (del[x.second]) continue;
      if (cnt[x.second] == k) {
        if ((flag = judge(x.second))) {
          printf("2\n");
          for (int y : vec[x.second])
            if (!del[y]) {
              printf("%d ", y + 1);
            }
          printf("\n");
          break;
        }
      }
      del[x.second] = true;
      for (auto it = vec[x.second].begin(); it != vec[x.second].end(); ++it)
        if (*it != x.second) {
          if (--cnt[*it] <= k && !del[*it]) {
            que.push(std::pair<int, int>(cnt[*it], *it));
          }
        }
    }
    if (flag) continue;
    int cnt2 = 0;
    for (int i = 0; i < n; ++i)
      if (!del[i]) ++cnt2;
    if (cnt2 == 0) {
      printf("-1\n");
      continue;
    }
    printf("1 %d\n", cnt2);
    for (int i = 0; i < n; ++i)
      if (!del[i]) {
        printf("%d ", i + 1);
      }
    printf("\n");
  }
  return 0;
}