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("O2")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int MAX = 1e5 + 5;
const long long MAX2 = 11;
const long long MOD = 1100000007;
const long long MOD2 = 1000005329;
const long long INF = 2e18;
const int dr[] = {1, 0, -1, 0, 1, 1, -1, -1, 0};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1, 0};
const double pi = acos(-1);
const double EPS = 1e-9;
const int block = 1000;
int tc, n, m, k, a, b, cnt[MAX];
vector<int> v[MAX], tmp;
unordered_set<int> u[MAX];
set<pair<int, int> > s;
inline bool make(int x) {
  s.clear();
  for (long long i = 1; i <= n; ++i) {
    u[i].clear();
    for (int j : v[i]) u[i].insert(j);
    cnt[i] = v[i].size();
    s.insert({v[i].size(), i});
  }
  while (s.size() > x && (*s.begin()).first < x) {
    a = (*s.begin()).second;
    s.erase(*s.begin());
    for (int i : u[a]) {
      s.erase({cnt[i], i});
      u[i].erase(a);
      --cnt[i];
      s.insert({cnt[i], i});
    }
  }
  if (s.size() == x) return false;
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> tc;
  for (long long ntc = 1; ntc <= tc; ++ntc) {
    cin >> n >> m >> k;
    for (long long i = 1; i <= n; ++i) v[i].clear();
    for (long long i = 1; i <= m; ++i) {
      cin >> a >> b;
      v[a].push_back(b);
      v[b].push_back(a);
    }
    if ((long long)k * (k - 1) / 2 > m) {
      cout << "-1\n";
      continue;
    }
    if (k == 1) {
      cout << "2\n1\n";
      continue;
    }
    if (k == 2) {
      if (n == 1) return 1;
      for (long long i = 1; i <= n; ++i)
        if (!v[i].empty()) {
          cout << "2\n" << i << ' ' << v[i][0] << '\n';
          break;
        }
      continue;
    }
    if (make(k)) {
      cout << "1 " << s.size() << '\n';
      for (pair<int, int> i : s) cout << i.second << ' ';
      cout << '\n';
    } else if (make(k - 1)) {
      bool st = 0;
      while (s.size() >= k) {
        if ((*s.begin()).first > k - 1) return 1;
        a = (*s.begin()).second, b = (*s.begin()).first;
        s.erase(*s.begin());
        if (b == k - 1) {
          tmp.clear();
          for (int i : u[a]) tmp.push_back(i);
          int sz = tmp.size() - 1;
          st = 1;
          for (long long i = 0; i <= sz; ++i)
            for (long long j = i + 1; j <= sz; ++j) {
              if (!st) break;
              if (!u[tmp[i]].count(tmp[j])) st = 0;
            }
          if (st) {
            cout << "2\n";
            cout << a << ' ';
            for (int i : tmp) cout << i << ' ';
            cout << '\n';
            break;
          }
        }
        for (int i : u[a]) {
          s.erase({cnt[i], i});
          u[i].erase(a);
          --cnt[i];
          s.insert({cnt[i], i});
        }
      }
      if (!st) cout << "-1\n";
    } else
      cout << "-1\n" << '\n';
  }
  return 0;
}