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;
struct nodS {
  int in;
  bool viz;
  vector<int> v;
};
vector<pair<int, int>> e;
vector<pair<int, int>> hmap[1000009];
int get_hash(long long x, long long y) {
  if (x > y) swap(x, y);
  return (x * (long long)200153 + y * (long long)600959) % 1000003;
}
void del() {
  for (auto it : e) {
    int val = get_hash(it.first, it.second);
    hmap[val].clear();
  }
}
void add(int x, int y) {
  if (x > y) swap(x, y);
  hmap[get_hash(x, y)].push_back({x, y});
}
bool find_hash(int x, int y) {
  if (x > y) swap(x, y);
  int val = get_hash(x, y);
  for (auto it : hmap[val])
    if (it.first == x && it.second == y) return true;
  return false;
}
bool check(vector<int> &v) {
  int sz = v.size();
  for (int i = 0; i < sz; i++)
    for (int j = i + 1; j < sz; j++) {
      if (!find_hash(v[i], v[j])) return false;
    }
  return true;
}
int n, m, k;
nodS nod[200005];
void solve() {
  del();
  e.clear();
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    nod[i].v.clear();
    nod[i].viz = false;
    nod[i].in = 0;
  }
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    nod[x].v.push_back(y);
    nod[y].v.push_back(x);
    nod[x].in++;
    nod[y].in++;
    add(x, y);
    e.push_back({x, y});
  }
  priority_queue<pair<long long, int>, vector<pair<long long, int>>,
                 greater<pair<long long, int>>>
      q;
  for (int i = 1; i <= n; i++) q.push({nod[i].in, i});
  while (!q.empty()) {
    int p = q.top().second;
    q.pop();
    if (nod[p].viz) continue;
    nod[p].viz = true;
    if (nod[p].in >= k) {
      vector<int> ans;
      for (int i = 1; i <= n; i++) {
        if (nod[i].in >= k) ans.push_back(i);
      }
      cout << "1 " << ans.size() << '\n';
      for (int it : ans) cout << it << ' ';
      cout << '\n';
      return;
    }
    if (nod[p].in == k - 1) {
      vector<int> v;
      v.push_back(p);
      for (int it : nod[p].v)
        if (!nod[it].viz) v.push_back(it);
      if (check(v)) {
        cout << "2\n";
        for (int it : v) cout << it << ' ';
        cout << '\n';
        return;
      }
    }
    for (int it : nod[p].v) {
      nod[it].in--;
      if (nod[it].in < k && !nod[it].viz) q.push({nod[it].in, it});
    }
  }
  cout << "-1\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t) {
    t--;
    solve();
  }
}