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;
long long n, m, k;
vector<vector<long long> > g;
vector<long long> num, u;
unordered_set<long long> S;
bool click(int v) {
  vector<long long> z;
  for (auto to : g[v])
    if (u[to]) z.push_back(to);
  for (int i = 0; i < z.size(); ++i) {
    for (int j = 0; j < i; ++j) {
      long long a = z[i];
      long long b = z[j];
      if (a > b) swap(a, b);
      if (!S.count((a << 20) ^ b)) return false;
    }
  }
  return true;
}
void f() {
  cin >> n >> m >> k;
  g.clear();
  g.resize(n);
  S.clear();
  for (int i = 0; i < m; ++i) {
    long long a, b;
    cin >> a >> b;
    --a;
    --b;
    g[a].push_back(b);
    g[b].push_back(a);
    if (a > b) swap(a, b);
    S.insert((a << 20) ^ b);
  }
  if (k >= 500) {
    cout << "-1\n";
    return;
  }
  num.resize(n);
  u.assign(n, 1);
  set<int> se;
  for (int i = 0; i < n; ++i) {
    num[i] = g[i].size();
    if (num[i] < k) {
      se.insert(i);
      u[i] = 0;
    }
  }
  while (!se.empty()) {
    int v = *se.begin();
    se.erase(se.begin());
    for (auto to : g[v]) {
      --num[to];
      if (!u[to]) continue;
      if (num[to] < k) {
        se.insert(to);
        u[to] = 0;
      }
    }
  }
  int al = accumulate(u.begin(), u.end(), 0);
  if (al > 0) {
    cout << 1 << " " << al << "\n";
    for (int i = 0; i < n; ++i)
      if (u[i]) cout << i + 1 << " ";
    cout << "\n";
    return;
  }
  u.assign(n, 1);
  se.clear();
  for (int i = 0; i < n; ++i) {
    num[i] = g[i].size();
    if (num[i] < k - 1) {
      se.insert(i);
      u[i] = 0;
    }
  }
  while (!se.empty()) {
    int v = *se.begin();
    se.erase(se.begin());
    for (auto to : g[v]) {
      --num[to];
      if (!u[to]) continue;
      if (num[to] < k - 1) {
        se.insert(to);
        u[to] = 0;
      }
    }
  }
  al = accumulate(u.begin(), u.end(), 0);
  if (al == 0) {
    cout << "-1\n";
    return;
  }
  for (int i = 0; i < n; ++i)
    if (num[i] == k - 1) se.insert(i);
  long long lim = (k * (k - 1)) / 2;
  while (!se.empty() && S.size() >= lim) {
    int v = *se.begin();
    se.erase(se.begin());
    if (num[v] == k - 1 && click(v)) {
      cout << 2 << "\n";
      cout << v + 1 << " ";
      for (auto to : g[v])
        if (u[to]) cout << to + 1 << " ";
      cout << "\n";
      return;
    }
    u[v] = 0;
    for (auto to : g[v]) {
      long long a = v;
      long long b = to;
      if (a > b) swap(a, b);
      S.erase((a << 20) ^ b);
      --num[to];
      if (!u[to]) continue;
      if (num[to] <= k - 1) se.insert(to);
    }
  }
  cout << "-1\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) f();
  return 0;
}