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;
using ll = long long;
using ld = long double;
template <class T>
bool ckmin(T& a, const T& b) {
  return a > b ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
const int N = 1 << 17;
vector<int> g[N], mn[N];
unordered_set<int> h[N];
int deg[N];
bool vis[N];
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  for (auto u = (0); u < (n); ++u) g[u].clear(), h[u].clear();
  for (auto e = (0); e < (m); ++e) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
    h[u].insert(v);
    h[v].insert(u);
  }
  if (1ll * k * (k - 1) / 2 > m) return void(cout << "-1\n");
  for (auto u = (0); u < (n); ++u) mn[u].clear();
  for (auto u = (0); u < (n); ++u) {
    vis[u] = 0;
    deg[u] = int((g[u]).size());
    mn[deg[u]].push_back(u);
  }
  for (int x = 0; x < n; x = mn[x].empty() ? x + 1 : x) {
    if (mn[x].empty()) continue;
    int u = mn[x].back();
    mn[x].pop_back();
    if (vis[u]) continue;
    vis[u] = true;
    g[u].erase(
        partition(begin(g[u]), end(g[u]), [&](int x) { return !vis[x]; }),
        end(g[u]));
    if (x == k - 1) {
      if ([&]() {
            for (int v : g[u])
              for (int x : g[u])
                if (x != v && !h[v].count(x)) return false;
            return true;
          }()) {
        cout << "2\n" << u + 1;
        for (int v : g[u]) cout << ' ' << v + 1;
        cout << '\n';
        return;
      }
    }
    if (x < k) {
      for (int v : g[u]) {
        --deg[v];
        mn[deg[v]].push_back(v);
        ckmin(x, deg[v]);
      }
      continue;
    }
    vector<int> res = {u};
    for (auto i = (0); i < (n); ++i)
      for (int v : mn[i])
        if (!vis[v]) res.push_back(v), vis[v] = true;
    cout << 1 << ' ' << int((res).size()) << '\n';
    for (int x : res) cout << x + 1 << ' ';
    cout << '\n';
    return;
  }
  cout << "-1\n";
}
signed main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) solve();
}