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("O3")
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1e5 + 5;
int n, m, k;
int deg[N], stat[N], val[N], rox[N];
vector<int> G[N];
set<int> intr, kozak;
vector<long long> g;
void zbij(int v) {
  deg[v]--;
  if (stat[v] == 2 && deg[v] <= k - 1) {
    stat[v]--;
    kozak.erase(v);
    intr.insert(v);
  }
  if (stat[v] == 1 && deg[v] <= k - 2) {
    stat[v]--;
    intr.erase(v);
    for (auto &x : (G[v]))
      if (stat[x]) rox[x] ^= val[v], zbij(x);
  }
}
void solve() {
  cin >> n >> m >> k;
  for (int i = (0); i < (n); i++) G[i].clear();
  g.clear();
  intr.clear();
  kozak.clear();
  for (int i = (0); i < (n); i++) val[i] = rox[i] = rand();
  for (int i = (0); i < (m); i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    G[a].push_back(b);
    G[b].push_back(a);
    rox[a] ^= val[b];
    rox[b] ^= val[a];
    g.push_back(1ll * INF * a + b);
    g.push_back(1ll * INF * b + a);
  }
  sort(g.begin(), g.end());
  for (int i = (0); i < (n); i++) {
    deg[i] = ((int)(G[i]).size()) + 1;
    stat[i] = 2;
    kozak.insert(i);
  }
  for (int i = (0); i < (n); i++) zbij(i);
  while (((int)(intr).size()) > 0) {
    int v = *intr.begin();
    vector<int> ziomy;
    for (auto &x : (G[v]))
      if (stat[x]) ziomy.push_back(x);
    bool ok = 1;
    for (auto &x : (ziomy))
      if (stat[x] == 1 && rox[x] != rox[v]) ok = 0;
    for (int i = (0); i < (((int)(ziomy).size())); i++) {
      if (!ok) break;
      for (int j = (i + 1); j < (((int)(ziomy).size())); j++) {
        if (!ok) break;
        long long co = 1ll * INF * ziomy[i] + ziomy[j];
        int ind = lower_bound(g.begin(), g.end(), co) - g.begin();
        if (ind == ((int)(g).size()))
          ok = 0;
        else
          ok &= g[ind] == co;
      }
    }
    if (ok) {
      cout << 2 << '\n';
      for (auto &x : (ziomy)) cout << x + 1 << ' ';
      cout << v + 1 << '\n';
      return;
    }
    zbij(v);
  }
  if (((int)(kozak).size()) > 0) {
    cout << 1 << ' ' << ((int)(kozak).size()) << '\n';
    for (auto &x : (kozak)) cout << x + 1 << ' ';
    cout << '\n';
  } else
    cout << -1 << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  srand(123);
  int tt;
  cin >> tt;
  for (int te = (0); te < (tt); te++) {
    solve();
  }
  return 0;
}