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;
double start_moment = 0;
double get_runtime() { return 1.0 * clock() / CLOCKS_PER_SEC; }
void reset_timer() { start_moment = get_runtime(); }
double timer_time() { return get_runtime() - start_moment; }
void runtime() { cout << fixed << setprecision(5) << get_runtime() << '\n'; }
template <class T>
void read(vector<T> &a, long long n) {
  T x;
  a.clear();
  for (long long i = 0; i < n; i++) {
    cin >> x;
    a.push_back(x);
  }
}
template <class T>
void write(vector<T> &a) {
  for (T x : a) cout << x << ' ';
  cout << '\n';
}
const int N = 100005;
int n, m, k;
set<int> g[N];
bool cmp(int a, int b) { return g[a].size() < g[b].size(); }
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    for (int i = 1; i <= n; i++) {
      g[i].clear();
    }
    scanf("%d %d %d", &n, &m, &k);
    set<pair<int, int> > e;
    for (int i = 0; i < m; i++) {
      int a, b;
      scanf("%d %d", &a, &b);
      g[a].insert(b);
      g[b].insert(a);
    }
    if (k > 500) {
      printf("-1\n");
      continue;
    }
    set<pair<int, int> > st;
    for (int i = 1; i <= n; i++) {
      st.insert({(int)g[i].size(), i});
    }
    while (1) {
      if (st.empty()) {
        printf("-1\n");
        break;
      }
      pair<int, int> p = *st.begin();
      if (p.first == k - 1) {
        vector<int> ans;
        for (int to : g[p.second]) {
          ans.push_back(to);
        }
        bool good = 1;
        if (ans.size() == k - 1) {
          sort(ans.begin(), ans.end(), &cmp);
          for (int i = 0; i < ans.size() && good; i++) {
            for (int j = i + 1; j < ans.size(); j++) {
              if (!g[ans[i]].count(ans[j])) {
                good = 0;
                break;
              }
            }
          }
        }
        ans.push_back(p.second);
        if (good) {
          printf("2\n");
          for (int x : ans) {
            printf("%d ", x);
          }
          printf("\n");
          break;
        }
      }
      if (p.first >= k) {
        printf("1 %d\n", (int)st.size());
        for (pair<int, int> p : st) {
          printf("%d ", p.second);
        }
        printf("\n");
        break;
      } else {
        st.erase(st.begin());
        for (int to : g[p.second]) {
          if (st.count({g[to].size(), to})) {
            st.erase({g[to].size(), to});
            g[to].erase(p.second);
            if (g[to].size()) {
              st.insert({g[to].size(), to});
            }
          }
        }
      }
    }
  }
}