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 nmax = 1e6 + 42;
int n, m, k;
set<int> adj[nmax];
int in[nmax];
queue<int> to_remove, idle;
vector<int> to_check[nmax];
bool there(int p, int q) {
  int pos = lower_bound(to_check[p].begin(), to_check[p].end(), q) -
            to_check[p].begin();
  if (pos < to_check[p].size() && to_check[p][pos] == q) return 1;
  return 0;
}
vector<pair<int, int> > busted;
bool check(int node, set<int> cur) {
  if (in[node] < k - 1) return 0;
  for (auto other : cur)
    if (in[other] < k - 1) return 0;
  cur.insert(node);
  for (auto p : busted)
    if (cur.count(p.first) && cur.count(p.second)) return 0;
  int in = 0;
  for (auto p : cur)
    for (auto q : cur)
      if (p < q) {
        if (there(p, q) == 0 || there(q, p) == 0) {
          if (in > 1000) busted.push_back({p, q});
          return 0;
        } else
          in += 2;
      }
  printf("2\n");
  for (auto k : cur) printf("%i ", k);
  printf("\n");
  return 1;
}
double T = clock();
void solve() {
  busted = {};
  scanf("%i%i%i", &n, &m, &k);
  for (int i = 1; i <= n; i++) adj[i] = {}, in[i] = 0, to_check[i] = {};
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%i%i", &u, &v);
    adj[u].insert(v);
    adj[v].insert(u);
    in[u]++;
    in[v]++;
    to_check[u].push_back(v);
    to_check[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) sort(to_check[i].begin(), to_check[i].end());
  to_remove = idle;
  for (int i = 1; i <= n; i++)
    if (in[i] < k) {
      to_remove.push(i);
    }
  while (to_remove.size()) {
    int tp = to_remove.front();
    if (in[tp] == k - 1) {
      if (check(tp, adj[tp])) return;
    }
    to_remove.pop();
    for (auto p : adj[tp]) {
      in[p]--;
      adj[p].erase(tp);
      if (in[p] == k - 1) to_remove.push(p);
    }
  }
  vector<int> outp = {};
  for (int i = 1; i <= n; i++)
    if (adj[i].size() >= k) outp.push_back(i);
  if (outp.size()) {
    printf("1\n");
    printf("%i", outp.size());
    for (auto w : outp) printf(" %i", w);
    printf("\n");
    return;
  }
  printf("-1\n");
}
int main() {
  int t;
  scanf("%i", &t);
  while (t) {
    t--;
    solve();
  }
  return 0;
}