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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
inline long long rand(long long x, long long y) {
  ++y;
  return (rng() % (y - x)) + x;
}
long long t, n, m, k, a, b, deg[200005];
bool keep[200005];
vector<long long> adj[200005];
void solve() {
  cin >> n >> m >> k;
  for (long long i = (long long)1; i <= (long long)n; i++) {
    adj[i].clear();
    keep[i] = 1;
    deg[i] = 0;
  }
  for (long long i = (long long)1; i <= (long long)m; i++) {
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
    deg[a]++, deg[b]++;
  }
  if (k * (k - 1) / 2 > m) {
    cout << "-1\n";
    return;
  }
  queue<long long> q;
  for (long long i = (long long)1; i <= (long long)n; i++) {
    sort(adj[i].begin(), adj[i].end());
    if (deg[i] < k - 1) q.push(i);
  }
  while (!q.empty()) {
    long long x = q.front();
    q.pop();
    if (!keep[x]) continue;
    keep[x] = 0;
    for (auto it : adj[x])
      if (keep[it]) {
        deg[it]--;
        if (deg[it] < k - 1) q.push(it);
      }
  }
  for (long long i = (long long)1; i <= (long long)n; i++)
    if (deg[i] == k - 1) q.push(i);
  while (!q.empty()) {
    long long x = q.front();
    q.pop();
    if (!keep[x]) continue;
    keep[x] = 0;
    if (deg[x] == k - 1) {
      bool ok = 1;
      vector<long long> v(1, x);
      for (auto it : adj[x])
        if (keep[it]) v.push_back(it);
      for (long long i = (long long)0; i <= (long long)k - 1; i++)
        for (long long j = (long long)i + 1; j <= (long long)k - 1; j++)
          if (!binary_search(adj[v[i]].begin(), adj[v[i]].end(), v[j])) ok = 0;
      if (ok) {
        cout << "2\n"
             << " ";
        for (auto it : v) cout << it << " ";
        cout << "\n";
        return;
      }
    }
    for (auto it : adj[x])
      if (keep[it]) {
        deg[it]--;
        if (deg[it] < k) q.push(it);
      }
  }
  vector<long long> v;
  for (long long i = (long long)1; i <= (long long)n; i++)
    if (keep[i]) v.push_back(i);
  if (v.empty())
    cout << "-1\n";
  else {
    cout << "1 " << (long long)v.size() << "\n";
    for (auto it : v) cout << it << " ";
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) solve();
}