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 main() {
  int t;
  cin >> t;
  while (t--) {
    int n, m, k;
    cin >> n >> m >> k;
    vector<vector<int> > g(n);
    vector<int> deg(n);
    for (int i = 0; i < m; i++) {
      int x, y;
      cin >> x >> y;
      --x;
      --y;
      g[x].push_back(y);
      g[y].push_back(x);
      deg[x]++;
      deg[y]++;
    }
    for (int i = 0; i < n; i++) sort(g[i].begin(), g[i].end());
    vector<bool> used(n), vis(n);
    vector<int> q(n);
    int top = 0;
    vector<int> vert(k);
    bool f = 0;
    for (int i = 0; i < n; i++) {
      used[i] = vis[i] = 0;
      if (deg[i] < k) {
        vis[i] = 1;
        q[top++] = i;
      }
    }
    for (int i = 0; i < top; i++) {
      int u = q[i];
      vis[u] = 0;
      used[u] = 1;
      if (deg[u] < k - 1) {
        for (auto j : g[u]) {
          if (!used[j]) --m;
          deg[j]--;
          if (!vis[j] and !used[j] and deg[j] < k) {
            vis[j] = 1;
            q[top++] = j;
          }
        }
        continue;
      } else if (deg[u] == k - 1) {
        if ((long long)k * (k - 1) <= 2 * (long long)m) {
          int r = 0;
          for (auto p : g[u]) {
            if (!used[p]) vert[r++] = p;
          }
          bool ff = 1;
          for (int x = 0; x < r; x++) {
            if (!ff) break;
            for (int y = x + 1; y < r; y++) {
              int pp =
                  lower_bound(g[vert[x]].begin(), g[vert[x]].end(), vert[y]) -
                  g[vert[x]].begin();
              if (pp == g[vert[x]].size() or g[vert[x]][pp] != vert[y]) {
                ff = 0;
                break;
              }
            }
          }
          if (ff) {
            f = 1;
            cout << "2\n" << u + 1 << ' ';
            for (int c = 0; c < r; c++) cout << vert[c] + 1 << ' ';
            cout << '\n';
            break;
          }
        }
      }
      for (auto j : g[u]) {
        deg[j]--;
        if (!used[j]) --m;
        if (!vis[j] and !used[j] and deg[j] < k) {
          vis[j] = 1;
          q[top++] = j;
        }
      }
    }
    if (f) continue;
    vector<int> ff;
    for (int i = 0; i < n; i++)
      if (!used[i]) ff.push_back(i);
    if (!ff.empty()) {
      cout << "1 " << ff.size() << '\n';
      for (auto c : ff) {
        c++;
        cout << c << ' ';
      }
      cout << '\n';
    } else
      cout << "-1\n";
  }
  return 0;
}