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>
#pragma GCC optimize("O2,unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2,ssse3,sse4,sse4.1")
using namespace std;
using ll = long long;
const int maxn = 1 << 18, mod = 1000000007;
inline __attribute__((always_inline, optimize("O2"))) int slower_bound(int *a,
                                                                       int n,
                                                                       int x) {
  int i = -1;
  i += 65536 * (i + 65536 < n && a[i + 65536] < x);
  i += 32768 * (i + 32768 < n && a[i + 32768] < x);
  i += 16384 * (i + 16384 < n && a[i + 16384] < x);
  i += 8192 * (i + 8192 < n && a[i + 8192] < x);
  i += 4096 * (i + 4096 < n && a[i + 4096] < x);
  i += 2048 * (i + 2048 < n && a[i + 2048] < x);
  i += 1024 * (i + 1024 < n && a[i + 1024] < x);
  i += 512 * (i + 512 < n && a[i + 512] < x);
  i += 256 * (i + 256 < n && a[i + 256] < x);
  i += 128 * (i + 128 < n && a[i + 128] < x);
  i += 64 * (i + 64 < n && a[i + 64] < x);
  i += 32 * (i + 32 < n && a[i + 32] < x);
  i += 16 * (i + 16 < n && a[i + 16] < x);
  i += 8 * (i + 8 < n && a[i + 8] < x);
  i += 4 * (i + 4 < n && a[i + 4] < x);
  i += 2 * (i + 2 < n && a[i + 2] < x);
  i += 1 * (i + 1 < n && a[i + 1] < x);
  return ++i < n && a[i] == x;
}
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> deg(n + 1), ded(n + 1);
  vector<vector<int>> g(n + 1);
  for (int f, t, i = 0; i < m; i++) {
    cin >> f >> t;
    g[f].push_back(t);
    g[t].push_back(f);
    deg[f]++, deg[t]++;
  }
  set<array<int, 2>> p;
  auto remove = [&](int v) {
    ded[v] = 1;
    p.erase({deg[v], v});
    for (auto &i : g[v])
      if (!ded[i]) {
        p.erase({deg[i], i});
        deg[i]--;
        p.insert({deg[i], i});
      }
  };
  for (int i = 1; i <= n; i++)
    p.insert({deg[i], i}), sort(begin(g[i]), end(g[i]));
  int C = 1;
  vector<int> lst(n + 1);
  while (!p.empty() && p.begin()->at(0) < k) {
    int v = p.begin()->at(1);
    C++;
    if (deg[v] == k - 1) {
      int ok = 1;
      vector<int> cli{v};
      lst[v] = C;
      for (auto i : g[v])
        if (!ded[i]) ok &= deg[i] >= k - 1, cli.push_back(i), lst[i] = C;
      if (ok) {
        int cur = 0;
        for (auto i : cli)
          for (auto j : cli)
            if (i < j) cur += slower_bound(g[i].data(), g[i].size(), j);
        if (cur < k * 1ll * (k - 1) / 2) {
          ok = 0;
        }
        if (ok) {
          cout << "2\n";
          for (auto i : cli) cout << i << " ";
          cout << '\n';
          return;
        }
      }
    }
    remove(p.begin()->at(1));
  }
  if (p.empty()) {
    cout << "-1\n";
    return;
  }
  vector<int> ans;
  for (int i = 1; i <= n; i++)
    if (!ded[i]) ans.push_back(i);
  cout << "1 " << ans.size() << '\n';
  for (auto &i : ans) cout << i << " ";
  cout << '\n';
}
int32_t main() {
  cin.tie(0)->sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}