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 T, n, m, k, x, y, dep[100005];
set<pair<int, int> > heap;
unordered_set<int> adj[100005];
int main() {
  scanf("%d", &T);
  for (; T--;) {
    scanf("%d %d %d", &n, &m, &k);
    heap.clear();
    for (int i = 1; i <= n; i++) adj[i].clear(), dep[i] = 0;
    for (int i = 1; i <= m; i++) {
      scanf("%d %d", &x, &y);
      dep[x]++;
      dep[y]++;
      adj[x].insert(y);
      adj[y].insert(x);
    }
    for (int i = 1; i <= n; i++) heap.insert(make_pair(dep[i], i));
    int flag = 0;
    for (; !heap.empty();) {
      pair<int, int> v = *heap.begin();
      if (v.first < k - 1) {
        for (auto p : adj[v.second]) {
          adj[p].erase(v.second);
          heap.erase(make_pair(dep[p], p));
          dep[p]--;
          heap.insert(make_pair(dep[p], p));
        }
        heap.erase(v);
      } else if (v.first >= k) {
        flag = 1;
        printf("1 %d\n", (int)heap.size());
        for (auto p : heap) printf("%d ", p.second);
        printf("\n");
        break;
      } else {
        if (k * (k - 1) <= 2 * m) {
          bool find = true;
          for (auto p = adj[v.second].begin(); find && p != adj[v.second].end();
               ++p)
            for (auto q = p; find && ++q != adj[v.second].end();)
              find &= (adj[*p].find(*q) != adj[*p].end());
          if (find) {
            printf("2\n");
            printf("%d ", v.second);
            for (int p : adj[v.second]) printf("%d ", p);
            printf("\n");
            flag = 1;
            break;
          }
        }
        for (auto p : adj[v.second]) {
          adj[p].erase(v.second);
          heap.erase(make_pair(dep[p], p));
          dep[p]--;
          heap.insert(make_pair(dep[p], p));
        }
        heap.erase(v);
      }
    }
    if (!flag) printf("-1\n");
  }
  return 0;
}