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("Ofast,no-stack-protector,unroll-loops")
#pragma GCC optimize("no-stack-protector,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int N = 1e5 + 100, OO = 1e9 + 7, T = 22, M = 1e9 + 7, P = 6151, SQ = 1300,
          lg = 22;
int mark[N], deg[N], ct[N];
bool ans[N], can[N];
vector<int> v[N], A;
vector<pair<int, int> > ch[N];
bool cmp(int x, int y) { return mark[x] < mark[y]; }
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  A.clear();
  for (int i = 0; i <= n; i++)
    v[i].clear(), ch[i].clear(), mark[i] = deg[i] = ct[i] = 0,
                                 ans[i] = can[i] = false;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  if (k > 500) {
    cout << -1 << endl;
    return;
  }
  set<pair<int, int> > st;
  for (int i = 1; i <= n; i++) st.insert({deg[i] = (int)v[i].size(), i});
  int cnt = 1;
  while ((int)st.size()) {
    pair<int, int> p = *st.begin();
    if (p.first >= k) {
      cout << 1 << " " << (int)st.size() << endl;
      for (auto u : st) cout << u.second << " ";
      cout << endl;
      return;
    } else {
      st.erase(p);
      mark[p.second] = cnt;
      for (auto u : v[p.second])
        if (!mark[u]) st.erase({deg[u], u}), st.insert({--deg[u], u});
      A.push_back(p.second);
    }
    cnt++;
  }
  for (auto i : A) {
    int nxt = 0;
    for (auto u : v[i])
      if (mark[u] > mark[i]) ct[nxt++] = u, can[u] = true;
    for (auto u : ch[i])
      if (!can[u.first]) ans[u.second] = false;
    for (int j = 0; j < nxt; j++) can[ct[j]] = false;
    if (nxt != k - 1) continue;
    ans[i] = true;
    sort(ct, ct + nxt, cmp);
    for (int j = 0; j < nxt; j++)
      for (int k = j + 1; k < nxt; k++) ch[ct[j]].push_back({ct[k], i});
  }
  for (auto i : A) {
    if (!ans[i]) continue;
    cout << 2 << endl;
    cout << i << " ";
    for (auto u : v[i])
      if (mark[u] > mark[i]) cout << u << " ";
    cout << endl;
    return;
  }
  cout << -1 << endl;
  return;
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}