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;
namespace CF1439B {
const int N = 100010, M = 100010, BarK = 450;
int n, m, k, d[N] = {}, del[N] = {};
set<int> snei[N];
vector<int> nei[N];
bool check(int now) {
  int l[BarK] = {};
  int t = 0;
  l[++t] = now;
  for (auto p : nei[now])
    if (!del[p]) l[++t] = p;
  if (t < k) return false;
  assert(t == k);
  for (int i = 1; i <= t; ++i)
    for (int j = 1; j < i; ++j)
      if (!snei[l[i]].count(l[j])) return false;
  return true;
}
int start() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= n; ++i) nei[i].clear();
    for (int i = 1, u, v; i <= m; ++i) {
      scanf("%d%d", &u, &v);
      nei[u].push_back(v);
      nei[v].push_back(u);
    }
    if (k == 1) {
      printf("2\n1\n");
      continue;
    }
    if (k >= BarK || k * (k - 1) / 2 > m) {
      puts("-1");
      continue;
    }
    for (int i = 1; i <= n; ++i) {
      d[i] = nei[i].size();
      del[i] = false;
    }
    queue<int> q;
    for (int i = 1; i <= n; ++i)
      if (d[i] < k - 1) {
        q.push(i);
        del[i] = true;
      }
    while (!q.empty()) {
      int now = q.front();
      q.pop();
      for (auto p : nei[now])
        if (--d[p] < k - 1 && !del[p]) {
          q.push(p);
          del[p] = true;
        }
    }
    int tot = 0;
    for (int i = 1; i <= n; ++i) snei[i].clear();
    for (int i = 1; i <= n; ++i) {
      if (del[i]) continue;
      ++tot;
      for (auto p : nei[i])
        if (!del[p]) snei[i].insert(p);
      if (d[i] == k - 1) q.push(i);
    }
    bool clique = false;
    while (!q.empty()) {
      int now = q.front();
      q.pop();
      if (check(now)) {
        clique = true;
        puts("2");
        printf("%d", now);
        for (auto p : nei[now])
          if (!del[p]) printf(" %d", p);
        puts("");
        break;
      } else {
        --tot;
        del[now] = true;
        for (auto p : nei[now])
          if (--d[p] == k - 1) q.push(p);
      }
    }
    if (tot) {
      if (clique) continue;
      printf("1 %d\n", tot);
      for (int i = 1; i <= n; ++i)
        if (!del[i]) printf("%d ", i);
      printf("\n");
    } else
      puts("-1");
  }
  return 0;
}
}  // namespace CF1439B
int main() { return CF1439B::start(); }