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 MAXN = 3e5 + 20;
unordered_set<int> V[MAXN];
set<pair<int, int> > st;
int n, m, k;
void solve() {
  pair<int, int> tmp;
  while (!st.empty()) {
    tmp = *st.begin();
    int u = tmp.second;
    int d = tmp.first;
    if (d >= k) {
      printf("%d %d\n", 1, st.size());
      for (auto x : st) printf("%d ", x.second);
      printf("\n");
      return;
    } else if (d == k - 1) {
      int ok = 1;
      for (auto x : V[u]) {
        for (auto y : V[u]) {
          if (y != x && !V[y].count(x)) {
            ok = 0;
            break;
          }
        }
        if (ok == 0) break;
      }
      if (ok) {
        printf("2\n");
        printf("%d ", u);
        for (auto x : V[u]) printf("%d ", x);
        printf("\n");
        return;
      }
    }
    st.erase(tmp);
    for (auto x : V[u]) {
      if (st.count(pair<int, int>(V[x].size(), x))) {
        st.erase(pair<int, int>(V[x].size(), x));
        V[x].erase(u);
        st.insert(pair<int, int>(V[x].size(), x));
      }
    }
  }
  printf("-1\n");
}
int main() {
  int test;
  scanf("%d", &test);
  while (test--) {
    scanf("%d%d%d", &n, &m, &k);
    st.clear();
    for (int i = 1; i <= n; i++) V[i].clear();
    int u, v;
    for (int i = 1; i <= m; i++) {
      scanf("%d%d", &u, &v);
      V[u].insert(v);
      V[v].insert(u);
    }
    for (int i = 1; i <= n; i++) st.insert(pair<int, int>(V[i].size(), i));
    solve();
  }
}