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 + 7;
vector<int> G[maxn], clique;
int d[maxn], n, vis[maxn], vis1[maxn];
void deal(int k) {
  queue<int> q;
  for (int i = 1; i <= n; i++) {
    if (d[i] < k) {
      q.push(i);
      vis[i] = 1;
    }
  }
  bool flag = false;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    vis[u] = 1;
    if (d[u] == k - 1) {
      clique.clear();
      clique.push_back(u);
      for (int i = 0; i < G[u].size(); i++) {
        int v = G[u][i];
        if (vis1[v]) continue;
        clique.push_back(v);
      }
      bool tem = false;
      for (int i = 0; i < clique.size(); i++) {
        for (int j = 0; j < clique.size(); j++) {
          if (clique[i] == clique[j]) break;
          if (!binary_search(G[clique[i]].begin(), G[clique[i]].end(),
                             clique[j])) {
            tem = true;
          }
        }
      }
      if (!tem) {
        flag = true;
        break;
      }
      clique.clear();
    }
    vis1[u] = 1;
    for (int i = 0; i < G[u].size(); i++) {
      int v = G[u][i];
      d[v]--;
      if (vis[v]) continue;
      if (d[v] < k) {
        vis[v] = 1;
        q.push(v);
      }
    }
  }
  if (flag) {
    printf("2\n");
    for (int i = 0; i < clique.size(); i++) printf("%d ", clique[i]);
  } else {
    int cnt = 0;
    for (int i = 1; i <= n; i++)
      if (!vis[i]) cnt++;
    if (cnt > 0) {
      printf("1 %d\n", cnt);
      for (int i = 1; i <= n; i++) {
        if (!vis[i]) printf("%d ", i);
      }
    } else
      printf("-1");
  }
  printf("\n");
}
void Init() {
  for (int i = 1; i <= n; i++) {
    G[i].clear();
    vis[i] = vis1[i] = 0;
    d[i] = 0;
  }
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int m, k;
    scanf("%d%d%d", &n, &m, &k);
    Init();
    for (int i = 1, u, v; i <= m; i++) {
      scanf("%d%d", &u, &v);
      G[u].push_back(v);
      G[v].push_back(u);
    }
    if (k * (k - 1) > 2 * m) {
      printf("-1\n");
      continue;
    }
    for (int i = 1; i <= n; i++) {
      d[i] = G[i].size();
      sort(G[i].begin(), G[i].end());
    }
    deal(k);
  }
}