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;
long double PI = 3.14159265359;
int inf = 1000000000000000007;
int mod = 1000000007;
int mod1 = 998244353;
const bool multi = 1;
vector<int> G[100007];
int deg[100007];
unordered_map<int, bool> is[100007];
bool odw[100007];
int n;
vector<int> check(int k) {
  memset(odw, 0, sizeof odw);
  deque<int> Q;
  for (int i = 1; i <= n; i++) {
    deg[i] = (int)(G[i]).size();
    if (deg[i] < k) {
      odw[i] = 1;
      Q.push_back(i);
    }
  }
  while (!Q.empty()) {
    int v = Q[0];
    Q.pop_front();
    for (auto u : G[v]) {
      deg[u]--;
      if (!odw[u] && deg[u] < k) {
        odw[u] = 1;
        Q.push_back(u);
      }
    }
  }
  vector<int> res;
  for (int i = 1; i <= n; i++)
    if (!odw[i]) res.push_back(i);
  return res;
}
bool is_clique(vector<int> xd, int k) {
  if ((int)(xd).size() != k) return 0;
  bool ok = 1;
  for (int i = 0; i < (int)(xd).size(); i++) {
    for (int j = i + 1; j < (int)(xd).size(); j++) {
      ok &= is[xd[i]][xd[j]];
    }
  }
  return ok;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tt;
  if (multi)
    cin >> tt;
  else
    tt = 1;
  while (tt--) {
    int m, k, a, b;
    cin >> n >> m >> k;
    for (int i = 0; i < m; i++) {
      cin >> a >> b;
      G[a].push_back(b);
      G[b].push_back(a);
      is[a][b] = 1;
      is[b][a] = 1;
    }
    vector<int> xd = check(k);
    if ((int)(xd).size() > 0) {
      cout << 1 << " " << (int)(xd).size() << '\n';
      for (auto x : xd) cout << x << " ";
      cout << '\n';
    } else {
      xd = check(k - 1);
      if ((int)(xd).size() > 0) {
        deque<int> Q;
        for (auto x : xd)
          if (deg[x] == k - 1) Q.push_back(x);
        bool ok = 0;
        while ((int)(Q).size() > 0) {
          int v = Q[0];
          odw[v] = 1;
          Q.pop_front();
          vector<int> c;
          c.push_back(v);
          for (auto u : G[v]) {
            if (odw[u]) continue;
            c.push_back(u);
            deg[u]--;
            if (deg[u] == k - 1) Q.push_back(u);
          }
          if (is_clique(c, k)) {
            cout << 2 << '\n';
            for (auto x : c) cout << x << " ";
            cout << '\n';
            ok = 1;
            break;
          }
        }
        if (!ok) cout << -1 << '\n';
      } else
        cout << -1 << '\n';
    }
    for (int i = 1; i <= n; i++) {
      G[i].clear();
      is[i].clear();
    }
  }
  return 0;
}