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;
inline bool isvowel(char ch) {
  ch = tolower(ch);
  return (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u');
}
inline bool isprime(int n) {
  if (n < 2 || (n % 2 == 0 && n != 2)) return false;
  for (int i = 3; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
class Union {
 private:
  vector<int> saizu, link;

 public:
  Union(int n) {
    saizu.assign(n, 1);
    link.resize(n);
    iota(link.begin(), link.end(), 0);
  }
  int find(int n) {
    if (link[n] == n) return n;
    return link[n] = find(link[n]);
  }
  int same(int a, int b) { return find(a) == find(b); }
  void unite(int a, int b) {
    if (same(a, b)) return;
    a = find(a);
    b = find(b);
    if (saizu[a] < saizu[b]) swap(a, b);
    saizu[a] += saizu[b];
    link[b] = a;
  }
  int getsize(int a) { return saizu[find(a)]; }
};
const int mod = 1e9 + 7;
long long mode(long long a) {
  a %= mod;
  if (a < 0) a += mod;
  return a;
}
long long subt(long long a, long long b) { return mode(mode(a) - mode(b)); }
long long add(long long a, long long b) { return mode(mode(a) + mode(b)); }
long long mult(long long a, long long b) { return mode(mode(a) * mode(b)); }
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = mult(res, a);
    a = mult(a, a);
    b >>= 1;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, m, k;
    cin >> n >> m >> k;
    vector<vector<int>> graph(n + 1);
    vector<int> used(n + 1), deg(n + 1);
    set<pair<int, int>> st;
    for (int i = 0; i < m; i++) {
      int x, y;
      cin >> x >> y;
      graph[x].push_back(y);
      graph[y].push_back(x);
      deg[x]++;
      deg[y]++;
    }
    for (int i = 1; i <= n; i++) {
      sort(graph[i].begin(), graph[i].end());
    }
    for (int i = 1; i <= n; i++) {
      st.insert(make_pair(deg[i], i));
    }
    int bigflag = 0;
    while (int((st).size())) {
      auto to = *st.begin();
      if (to.first >= k) break;
      if (to.first == k - 1) {
        int flag = 1;
        vector<int> vec;
        vec.push_back(to.second);
        for (auto u : graph[to.second])
          if (!used[u]) vec.push_back(u);
        if (int((vec).size()) == k) {
          for (int i = 0; i < int((vec).size()); i++) {
            for (int j = i + 1; j < int((vec).size()); j++) {
              auto u = vec[i], v = vec[j];
              if (!binary_search(graph[u].begin(), graph[u].end(), v)) {
                flag = 0;
                break;
              }
            }
            if (!flag) break;
          }
          if (flag) {
            cout << 2 << "\n";
            for (auto u : vec) cout << u << ' ';
            cout << "\n";
            bigflag++;
          }
        }
      }
      if (bigflag) break;
      used[to.second]++;
      st.erase(st.begin());
      for (auto v : graph[to.second]) {
        if (!used[v]) {
          st.erase(st.find(make_pair(deg[v], v)));
          deg[v]--;
          st.insert(make_pair(deg[v], v));
        }
      }
    }
    if (bigflag) continue;
    if (!int((st).size())) {
      cout << -1;
    } else {
      cout << 1 << ' ' << int((st).size()) << "\n";
      for (auto to : st) cout << to.second << ' ';
    }
    cout << "\n";
  }
  return 0;
}