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;
const int Maxn = 100005;
const int Maxm = 500;
int T;
int n, m, k;
vector<int> un;
int my[Maxn];
bool has[Maxm][Maxm];
set<int> neigh[Maxn];
void Solve(priority_queue<pair<int, int> > &Q) {
  bool assigned = false;
  int inside = n;
  while (!Q.empty() && -Q.top().first < k) {
    int v = Q.top().second, d = -Q.top().first;
    Q.pop();
    if (neigh[v].size() != d) continue;
    if (neigh[v].size() == k - 1) {
      if (!assigned && inside < Maxm) {
        int pnt = 0;
        for (int i = 1; i <= n; i++)
          if (!neigh[i].empty()) my[i] = pnt++;
        for (int i = 0; i < pnt; i++) fill(has[i], has[i] + pnt, false);
        for (int i = 1; i <= n; i++)
          for (auto u : neigh[i]) has[my[i]][my[u]] = true;
        assigned = true;
      }
      vector<int> seq;
      for (auto u : neigh[v]) seq.push_back(u);
      bool ok = true;
      for (int i = 0; i < seq.size() && ok; i++)
        for (int j = i + 1; j < seq.size() && ok; j++)
          if (assigned)
            ok = has[my[seq[i]]][my[seq[j]]];
          else
            ok = neigh[seq[i]].find(seq[j]) != neigh[seq[i]].end();
      if (ok) {
        printf("2\n");
        printf("%d", v);
        for (int i = 0; i < seq.size(); i++) printf(" %d", seq[i]);
        printf("\n");
        return;
      }
    }
    for (auto u : neigh[v]) {
      neigh[u].erase(v);
      if (assigned) has[my[v]][my[u]] = has[my[u]][my[v]] = false;
      Q.push(pair<int, int>(-int(neigh[u].size()), u));
    }
    neigh[v].clear();
    inside--;
  }
  vector<int> seq;
  while (!Q.empty()) {
    int v = Q.top().second, d = -Q.top().first;
    Q.pop();
    if (neigh[v].size() != d) continue;
    seq.push_back(v);
  }
  if (seq.empty())
    printf("-1\n");
  else {
    printf("1 %d\n", int(seq.size()));
    for (int i = 0; i < seq.size(); i++)
      printf("%d%c", seq[i], i + 1 < seq.size() ? ' ' : '\n');
  }
}
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d %d %d", &n, &m, &k);
    for (int i = 1; i <= n; i++) neigh[i].clear();
    for (int i = 0; i < m; i++) {
      int a, b;
      scanf("%d %d", &a, &b);
      neigh[a].insert(b);
      neigh[b].insert(a);
    }
    priority_queue<pair<int, int> > Q;
    for (int i = 1; i <= n; i++)
      Q.push(pair<int, int>(-int(neigh[i].size()), i));
    Solve(Q);
  }
  return 0;
}