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 <typename F, typename... Args>
void timeit(F func, Args &&...args) {
  clock_t start = clock();
  func(std::forward<Args>(args)...);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int maxn = 1e5 + 100;
vector<int> G[maxn];
vector<pair<int, int>> check[maxn];
int deg[maxn], dfn[maxn], que[maxn];
bool ans[maxn], can[maxn];
void solve() {
  int n, m, k, cnt = 0;
  cin >> n >> m >> k;
  for (int i = 0; i <= n; i++) G[i].clear();
  for (int i = 0; i <= n; i++) check[i].clear();
  for (int i = 0; i <= n; i++) deg[i] = dfn[i] = 0;
  for (int i = 0; i <= n; i++) ans[i] = false;
  for (int i = 0, u, v; i < m; i++) {
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
    deg[u]++;
    deg[v]++;
  }
  if (1ll * k * (k - 1) > m * 2ll) {
    cout << "-1\n";
    return;
  }
  set<pair<int, int>> st;
  vector<int> pts;
  for (int i = 1; i <= n; i++) st.insert({deg[i], i});
  while (!st.empty()) {
    pair<int, int> tmp = *st.begin();
    if (tmp.first >= k) break;
    st.erase(st.begin());
    for (int v : G[tmp.second]) {
      if (dfn[v] == 0) {
        st.erase({deg[v], v});
        st.insert({--deg[v], v});
      }
    }
    dfn[tmp.second] = ++cnt;
    pts.push_back(tmp.second);
  }
  if (st.size()) {
    cout << 1 << ' ' << st.size() << '\n';
    for (pair<int, int> p : st) cout << p.second << ' ';
    cout << '\n';
    return;
  }
  for (int u : pts) {
    int len = 0;
    for (int v : G[u]) {
      if (dfn[v] < dfn[u]) continue;
      que[len++] = v;
      can[v] = true;
    }
    for (pair<int, int> p : check[u]) {
      if (!can[p.first]) ans[p.second] = false;
    }
    for (int i = 0; i < len; i++) can[que[i]] = false;
    if (len != k - 1) continue;
    ans[u] = true;
    sort(que, que + len, [&](int a, int b) { return dfn[a] < dfn[b]; });
    for (int i = 0; i < len; i++) {
      for (int j = i + 1; j < len; j++) {
        check[que[i]].push_back({que[j], u});
      }
    }
  }
  for (int u : pts) {
    if (!ans[u]) continue;
    cout << "2\n" << u << " ";
    for (int v : G[u]) {
      if (dfn[v] > dfn[u]) cout << v << " ";
    }
    cout << '\n';
    return;
  }
  cout << "-1\n";
}
signed main() {
  ios::sync_with_stdio(false), cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
}