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;
struct myHash {
  int operator()(int x) const { return x ^ (x << 1); }
};
unordered_set<int, myHash> gph[101010];
int deg[101010];
bool vst[101010];
queue<int> Q;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T;
  cin >> T;
  while (T--) {
    int n, m, k;
    cin >> n >> m >> k;
    for (int i = 0; i < n; ++i)
      unordered_set<int, myHash>().swap(gph[i]), deg[i] = 0, vst[i] = false;
    while (Q.size()) Q.pop();
    for (int i = 0; i < m; ++i) {
      int x, y;
      cin >> x >> y;
      --x;
      --y;
      gph[x].insert(y);
      ++deg[x];
      gph[y].insert(x);
      ++deg[y];
    }
    for (int i = 0; i < n; ++i)
      if (deg[i] < k - 1) Q.push(i);
    while (Q.size()) {
      int x = Q.front();
      Q.pop();
      deg[x] = 0;
      vst[x] = true;
      for (auto y : gph[x])
        if (!vst[y]) {
          if (--deg[y] == k - 2) {
            Q.push(y);
          }
        }
    }
    for (int i = 0; i < n; ++i)
      if (deg[i] == k - 1) Q.push(i);
    bool bigflag = true;
    while (Q.size()) {
      int x = Q.front();
      Q.pop();
      deg[x] = 0;
      vst[x] = true;
      vector<int> V;
      for (int y : gph[x])
        if (!vst[y]) V.push_back(y);
      bool flag = ((int)V.size() == k - 1);
      if (flag) {
        for (int i : V) {
          for (int j : V) {
            if (i != j && gph[i].count(j) == 0) {
              flag = false;
              break;
            }
          }
          if (!flag) break;
        }
      }
      if (flag) {
        cout << "2\n" << x + 1 << ' ';
        for (auto y : V) cout << y + 1 << ' ';
        cout << '\n';
        bigflag = false;
        break;
      }
      for (auto y : gph[x])
        if (!vst[y]) {
          if (--deg[y] == k - 1) {
            Q.push(y);
          }
        }
    }
    if (bigflag) {
      int cnt = 0;
      for (int i = 0; i < n; ++i)
        if (!vst[i]) ++cnt;
      if (cnt == 0)
        cout << "-1\n";
      else {
        cout << "1 " << cnt << '\n';
        for (int i = 0; i < n; ++i)
          if (!vst[i]) cout << i + 1 << ' ';
        cout << '\n';
      }
    }
  }
}