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;
int t;
int n, m, k;
vector<int> v[100086];
int x, y;
int deg[100086];
queue<int> q;
bool inq[100086];
vector<int> ans;
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= n; i++) v[i].clear(), deg[i] = inq[i] = 0;
    for (int i = 1; i <= m; i++) {
      scanf("%d%d", &x, &y);
      v[x].push_back(y), v[y].push_back(x);
      deg[x]++, deg[y]++;
    }
    if (1ll * k * (k - 1) / 2 > m) {
      puts("-1");
      continue;
    }
    for (int i = 1; i <= n; i++) sort(v[i].begin(), v[i].end());
    while (!q.empty()) q.pop();
    for (int i = 1; i <= n; i++)
      if (deg[i] <= k - 1) q.push(i), inq[i] = true;
    bool suc = false;
    while (!q.empty()) {
      int x = q.front();
      q.pop();
      if (deg[x] == k - 1) {
        vector<int> w;
        w.push_back(x);
        for (int i = 0; i < v[x].size(); i++) {
          int to = v[x][i];
          if (!deg[to]) continue;
          w.push_back(to);
        }
        bool tag = true;
        for (int i = 0; i < w.size() && tag; i++) {
          for (int j = 0; j < i && tag; j++) {
            tag &= binary_search(v[w[i]].begin(), v[w[i]].end(), w[j]);
          }
        }
        if (tag) {
          suc = true;
          puts("2");
          for (int i = 0; i < w.size(); i++) printf("%d ", w[i]);
          puts("");
          break;
        }
      }
      for (int i = 0; i < v[x].size(); i++) {
        int to = v[x][i];
        if (!deg[to]) continue;
        deg[to]--;
        if (deg[to] <= k - 1 && !inq[to]) q.push(to), inq[to] = true;
      }
      deg[x] = 0;
    }
    if (suc) continue;
    ans.clear();
    for (int i = 1; i <= n; i++)
      if (deg[i]) ans.push_back(i);
    if (ans.empty())
      puts("-1");
    else {
      printf("1 %d\n", ans.size());
      for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
      puts("");
    }
  }
}