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 maxn = 1e5 + 5;
vector<int> vv[maxn], uu[maxn], pp[maxn];
int degree[maxn];
int n, m, k;
queue<int> q;
int vis[maxn];
bool res[maxn];
int r;
void init() {
  r = 0;
  while (q.size()) q.pop();
  for (int i = 1; i <= n; i++) vv[i].clear(), pp[i].clear(), degree[i] = 0;
}
bool step1() {
  int cnt = n;
  for (int i = 1; i <= n; i++) {
    vis[i] = 0;
    if (degree[i] < k) {
      q.push(i);
    }
  }
  while (q.size()) {
    int u = q.front();
    vis[u] = 1;
    cnt--;
    q.pop();
    if (degree[u] == k - 1) {
      ++r;
      uu[r].clear();
      uu[r].push_back(u);
      pp[u].push_back(r);
      for (int i = 0; i < vv[u].size(); i++) {
        if (!vis[vv[u][i]]) {
          uu[r].push_back(vv[u][i]);
          pp[vv[u][i]].push_back(r);
        }
      }
    }
    for (int i = 0; i < vv[u].size(); i++) {
      int v = vv[u][i];
      if (vis[v] == 0) {
        degree[v]--;
        if (degree[v] == k - 1) q.push(v);
      }
    }
  }
  if (cnt > 0) {
    cout << 1 << " " << cnt << "\n";
    for (int i = 1; i <= n; i++) {
      if (degree[i] >= k) cout << i << " ";
    }
    cout << "\n";
    return 1;
  }
  return 0;
}
bool step2() {
  for (int i = 1; i <= n; i++) vis[i] = 0;
  for (int i = 1; i <= r; i++) res[i] = true;
  for (int i = 1; i <= n; i++) {
    if (pp[i].size() == 0) continue;
    vis[i] = 1;
    for (int j = 0; j < vv[i].size(); j++) vis[vv[i][j]] = 1;
    for (int j = 0; j < pp[i].size(); j++) {
      int tuan = pp[i][j];
      if (res[tuan]) {
        for (int k = 0; k < uu[tuan].size(); k++) {
          if (vis[uu[tuan][k]] == 0) {
            res[tuan] = false;
            break;
          }
        }
      }
    }
    vis[i] = 0;
    for (int j = 0; j < vv[i].size(); j++) vis[vv[i][j]] = 0;
  }
  for (int i = 1; i <= r; i++) {
    if (res[i]) {
      cout << 2 << "\n";
      for (int j = 0; j < uu[i].size(); j++) cout << uu[i][j] << " ";
      cout << "\n";
      return 1;
    }
  }
  return 0;
}
void solve() {
  cin >> n >> m >> k;
  init();
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    vv[x].push_back(y);
    vv[y].push_back(x);
    degree[x]++;
    degree[y]++;
  }
  if (k * (k - 1) > 2 * m) {
    cout << -1 << "\n";
    return;
  }
  if (step1()) {
    return;
  }
  if (step2()) {
    return;
  }
  cout << -1 << "\n";
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int _ = 1;
  cin >> _;
  while (_--) solve();
}