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;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <class T>
T sqr(T x) {
  return x * x;
}
const double eps = 1e-8;
const double pi = acos(-1.);
const int inf = 1e9 + 7;
const int N = 1e5 + 20;
int n, m, k;
vector<int> g[N];
int deg[N];
unordered_set<long long> w[N];
void solve() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) {
    deg[i] = 0;
    g[i].clear();
    w[i].clear();
  }
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    g[x].push_back(y);
    g[y].push_back(x);
    w[x].insert(y);
    w[y].insert(x);
    deg[x]++, deg[y]++;
  }
  if ((long long)k * (k - 1) > 2 * m) {
    cout << -1 << '\n';
    return;
  }
  auto cmp = [](int a, int b) {
    return deg[a] == deg[b] ? a < b : deg[a] < deg[b];
  };
  set<int, decltype(cmp)> cur(cmp);
  vector<vector<int> > check;
  for (int i = 0; i < n; i++) cur.insert(i);
  while (((int)cur.size())) {
    auto f = *cur.begin();
    if (deg[f] >= k) {
      cout << 1 << " " << ((int)cur.size()) << '\n';
      for (auto x : cur) cout << x + 1 << " ";
      cout << '\n';
      return;
    }
    cur.erase(f);
    if (deg[f] == k - 1) {
      vector<int> t = {f};
      for (auto x : g[f])
        if (cur.count(x)) t.push_back(x);
      check.push_back(t);
    }
    for (auto x : g[f])
      if (cur.count(x)) {
        cur.erase(x);
        deg[x]--;
        cur.insert(x);
      }
  }
  for (int i = 0; i < ((int)check.size()); i++) {
    auto& a = check[i];
    bool ok = 1;
    for (int j = 0; j < ((int)a.size()) && ok; j++)
      for (int p = j + 1; p < ((int)a.size()) && ok; p++)
        if (!w[a[j]].count(a[p])) {
          ok = 0;
        }
    if (ok) {
      cout << "2\n";
      for (auto x : a) cout << x + 1 << " ";
      cout << '\n';
      return;
    }
  }
  cout << -1 << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}