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 T1, typename T2>
inline void chkmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void chkmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
using ll = long long;
using ld = long double;
const string FILENAME = "input";
const int MAXN = 100228;
int n, m, k;
vector<int> g[MAXN];
int deg[MAXN];
bool del[MAXN];
bool in[MAXN];
map<int, bool> kek[MAXN];
int degs[MAXN];
bool check(vector<int> x) {
  for (auto y : x) {
    if ((int)(g[y]).size() < (int)(x).size() - 1) {
      return false;
    }
  }
  vector<int> d;
  for (int i = 0; i < (int)(x).size(); i++) {
    d.push_back(degs[x[i]]);
  }
  sort((d).begin(), (d).end());
  for (int i = 0; i < (int)(x).size(); i++) {
    if (d[i] < i) {
      return false;
    }
  }
  for (int i = 0; i < (int)(x).size(); i++) {
    for (int j = i + 1; j < (int)(x).size(); j++) {
      if (kek[x[i]].find(x[j]) == kek[x[i]].end()) {
        return false;
      }
    }
  }
  return true;
}
void solve() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) {
    g[i].clear();
    kek[i].clear();
  }
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    kek[a][b] = true;
    kek[b][a] = true;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  set<pair<int, int> > s;
  for (int i = 0; i < n; i++) {
    deg[i] = (int)(g[i]).size();
    del[i] = false;
    s.insert(make_pair(deg[i], i));
    in[i] = false;
  }
  vector<int> order;
  while (!s.empty()) {
    auto x = *s.begin();
    if (x.first >= k) {
      break;
    }
    s.erase(x);
    order.push_back(x.second);
    del[x.second] = true;
    for (auto y : g[x.second]) {
      if (!del[y]) {
        s.erase(make_pair(deg[y], y));
        deg[y]--;
        s.insert(make_pair(deg[y], y));
      }
    }
  }
  if (!s.empty()) {
    cout << 1 << ' ' << (int)(s).size() << '\n';
    for (auto h : s) {
      cout << h.second + 1 << ' ';
    }
    cout << '\n';
    return;
  }
  reverse((order).begin(), (order).end());
  for (int i = 0; i < (int)(order).size(); i++) {
    degs[order[i]] = 0;
    vector<int> kek;
    for (auto h : g[order[i]]) {
      if (in[h]) {
        degs[order[i]]++;
        kek.push_back(h);
      }
    }
    kek.push_back(order[i]);
    if ((int)(kek).size() == k) {
      if (check(kek)) {
        cout << 2 << '\n';
        for (auto y : kek) {
          cout << y + 1 << ' ';
        }
        cout << '\n';
        return;
      }
    }
    in[order[i]] = true;
  }
  cout << -1 << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}