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 inf = 0x3f3f3f3f;
const double eps = 1e-6;
const int maxn = 1e5 + 2000;
int n, m, k, u, v;
vector<int> g[maxn], clique;
queue<int> q;
int d[maxn], vis[maxn];
void init() {
  for (int i = 1; i <= n; i++) g[i].clear(), d[i] = 0, vis[i] = 0;
  while (!q.empty()) q.pop();
  clique.clear();
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    cin >> n >> m >> k;
    init();
    for (int i = 1; i <= m; i++)
      scanf("%d%d", &u, &v), g[u].push_back(v), g[v].push_back(u);
    for (int i = 1; i <= n; i++)
      sort(g[i].begin(), g[i].end()), d[i] = (int)g[i].size();
    for (int i = 1; i <= n; i++)
      if (d[i] < k) vis[i] = 1, q.push(i);
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      vis[u] = 2;
      if (d[u] == k - 1 && clique.empty()) {
        clique.push_back(u);
        for (auto v : g[u])
          if (vis[v] != 2) clique.push_back(v);
        bool flag = 1;
        for (auto x : clique)
          for (auto y : clique)
            if (x != y && !binary_search(g[x].begin(), g[x].end(), y)) flag = 0;
        if (!flag) clique.clear();
      }
      for (auto v : g[u]) {
        d[v]--;
        if (!vis[v] && d[v] < k) vis[v] = 1, q.push(v);
      }
    }
    int ans = 0;
    for (int i = 1; i <= n; i++)
      if (!vis[i]) ans++;
    if (ans) {
      printf("1 %d\n", ans);
      for (int i = 1; i <= n; i++)
        if (!vis[i]) printf("%d ", i);
      printf("\n");
    } else if (!clique.empty()) {
      printf("2\n");
      for (auto x : clique) printf("%d ", x);
      printf("\n");
    } else
      printf("-1\n");
  }
}