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 mxn = 2e5 + 5;
vector<int> adj[mxn];
int bad[mxn];
int vis[mxn];
int deg[mxn];
void solve() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    adj[i].clear();
    deg[i] = 0;
    vis[i] = 0;
    bad[i] = 0;
  }
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
    deg[u]++;
    deg[v]++;
  }
  for (int i = 1; i <= n; i++) {
    sort(adj[i].begin(), adj[i].end());
  }
  priority_queue<pair<int, int>, vector<pair<int, int>>,
                 greater<pair<int, int>>>
      pq;
  for (int i = 1; i <= n; i++) {
    pq.push({deg[i], i});
  }
  vector<int> ans;
  int clique = -1;
  while (!pq.empty()) {
    int u = pq.top().second;
    int d = pq.top().first;
    pq.pop();
    if (vis[u]) continue;
    vis[u] = 1;
    if (d >= k) {
      ans.push_back(u);
      continue;
    }
    if (d == k - 1 && (k * (k - 1)) / 2 <= m) {
      int ok = true;
      for (int i = 0; i < adj[u].size(); i++) {
        int x = adj[u][i];
        if (bad[x]) continue;
        for (int r = i + 1; r < adj[u].size(); r++) {
          int y = adj[u][r];
          if (bad[y]) continue;
          int f = binary_search(adj[x].begin(), adj[x].end(), y);
          if (!f) {
            ok = false;
            break;
          }
        }
        if (!ok) break;
      }
      if (ok) {
        clique = u;
        break;
      }
    }
    bad[u] = 1;
    for (auto v : adj[u]) {
      deg[v]--;
      pq.push({deg[v], v});
    }
  }
  if (clique != -1) {
    printf("2\n%d ", clique);
    for (int i : adj[clique]) {
      if (bad[i]) continue;
      printf("%d ", i);
    }
    printf("\n");
  } else if (ans.size() > k) {
    printf("1 %d\n", ans.size());
    for (int i : ans) {
      printf("%d ", i);
    }
    printf("\n");
    return;
  } else {
    printf("-1\n");
  }
  return;
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    solve();
  }
  return 0;
}