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")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <class T>
inline bool setmin(T &a, T b) {
  if (a > b) return a = b, 1;
  return 0;
}
template <class T>
inline bool setmax(T &a, T b) {
  if (a < b) return a = b, 1;
  return 0;
}
template <class T>
inline T fast(T a, T b, T mod) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
template <class T>
istream &operator>>(istream &os, vector<T> &container) {
  for (auto &u : container) os >> u;
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &container) {
  for (auto &u : container) os << u << " ";
  return os;
}
template <class T>
inline T gcd(T a, T b) {
  while (b) swap(a %= b, b);
  return abs(a);
}
const int INF = 1e9 + 7;
const long long mod = 998244353;
const long long BIG_INF = 1e18 + 7;
const long long N = 1e6 + 7;
const unsigned int T = 1LL << 20;
const long double inf = 1e18;
const long double eps = 1e-14;
const int LOG = 20;
long long n, m, k;
vector<vector<int> > G(N);
vector<unordered_set<int> > siema(N);
int stopien[N];
set<pair<int, int> > wziete;
bool check_clique(int v) {
  vector<int> teraz = {v};
  for (auto &u : G[v])
    if (wziete.count({stopien[u], u})) teraz.push_back(u);
  assert((int)(teraz).size() == k);
  shuffle((teraz).begin(), (teraz).end(), rng);
  random_shuffle((teraz).begin(), (teraz).end());
  for (int i = 0; i < k; i++)
    for (int j = i + 1; j < k; j++)
      if (!siema[teraz[i]].count(teraz[j])) return false;
  cout << 2 << '\n';
  for (auto &u : teraz) cout << u << ' ';
  cout << '\n';
  return true;
}
void solve() {
  cin >> n >> m >> k;
  srand(time(0));
  for (int i = 1; i <= n; i++) G[i].clear();
  for (int i = 1; i <= n; i++) siema[i].clear();
  wziete = set<pair<int, int> >();
  for (int i = 1; i <= n; i++) stopien[i] = 1;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
    siema[a].insert(b);
    siema[b].insert(a);
  }
  for (int i = 1; i <= n; i++) stopien[i] = (int)(G[i]).size();
  for (int i = 1; i <= n; i++) wziete.insert({stopien[i], i});
  while ((int)(wziete).size()) {
    if ((*wziete.begin()).first == k - 1)
      if (check_clique((*wziete.begin()).second)) {
        return;
      }
    if ((*wziete.begin()).first >= k) break;
    int pom = (*wziete.begin()).second;
    wziete.erase(wziete.begin());
    for (auto &u : G[pom]) {
      if (wziete.count({stopien[u], u})) {
        wziete.erase({stopien[u], u});
        stopien[u]--;
        wziete.insert(make_pair(stopien[u], u));
      }
    }
  }
  if ((int)(wziete).size() > k) {
    cout << 1 << ' ' << (int)(wziete).size() << '\n';
    for (auto &u : wziete) cout << u.second << ' ';
    cout << '\n';
  } else {
    cout << -1 << '\n';
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) solve();
}