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", "unroll-loops")
#pragma GCC target("sse4.2")
using namespace std;
mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
void flush() { cout << flush; }
void flushln() { cout << endl; }
void println() { cout << '\n'; }
template <class T>
void print(const T &x) {
  cout << x;
}
template <class T>
void read(T &x) {
  cin >> x;
}
template <class T, class... U>
void read(T &x, U &...u) {
  read(x);
  read(u...);
}
template <class T, class... U>
void print(const T &x, const U &...u) {
  print(x);
  print(u...);
}
template <class T, class... U>
void println(const T &x, const U &...u) {
  print(x);
  println(u...);
}
template <class T, class U>
inline bool chmin(T &x, const U &y) {
  return y < x ? x = y, 1 : 0;
}
template <class T, class U>
inline bool chmax(T &x, const U &y) {
  return y > x ? x = y, 1 : 0;
}
template <class T>
inline void sort(T &a) {
  sort((a).begin(), (a).end());
}
template <class T>
inline void rsort(T &a) {
  sort((a).rbegin(), (a).rend());
}
template <class T>
inline void reverse(T &a) {
  reverse((a).begin(), (a).end());
}
template <class T, class U>
inline istream &operator>>(istream &str, pair<T, U> &p) {
  return str >> p.first >> p.second;
}
template <class T>
inline istream &operator>>(istream &str, vector<T> &a) {
  for (auto &i : a) str >> i;
  return str;
}
template <class T>
inline T sorted(T a) {
  sort(a);
  return a;
}
void solve() {
  int n, m, k;
  read(n, m, k);
  vector<unordered_set<int>> G(n);
  for (int i = 0; i < m; ++i) {
    int u, v;
    read(u, v);
    --u, --v;
    G[u].insert(v);
    G[v].insert(u);
  }
  if (k > 2 * sqrt(m)) {
    println(-1);
    return;
  }
  set<pair<int, int>> currsz;
  for (int i = 0; i < n; ++i) {
    currsz.insert({signed((G[i]).size()), i});
  }
  while (signed((currsz).size())) {
    auto first = *currsz.begin();
    if (first.first >= k) break;
    currsz.erase(first);
    vector<int> v((G[first.second]).begin(), (G[first.second]).end());
    v.push_back(first.second);
    if (signed((v).size()) == k) {
      for (int i = 0; i < signed((v).size()); ++i)
        for (int j = i + 1; j < signed((v).size()); ++j)
          if (!G[v[i]].count(v[j])) goto go;
      println(2);
      for (auto i : v) print(i + 1, ' ');
      println();
      return;
    go:;
    }
    for (auto i : G[first.second]) {
      if (currsz.count({signed((G[i]).size()), i})) {
        currsz.erase({signed((G[i]).size()), i});
        G[i].erase(first.second);
        currsz.insert({signed((G[i]).size()), i});
      }
    }
  }
  if (signed((currsz).size())) {
    println(1, ' ', signed((currsz).size()));
    for (auto i : currsz) print(i.second + 1, ' ');
    println();
  } else {
    println(-1);
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  int t;
  read(t);
  while (t--) solve();
  return 0;
}