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 = 2e5 + 100;
int T;
int n, m, k;
int deg[maxn], vis[maxn], del[maxn], now;
vector<int> g[maxn], clique;
queue<int> q;
void init() {
  now = n;
  for (int i = 1; i <= n; i++) g[i].clear(), deg[i] = 0, vis[i] = 0, del[i] = 0;
  while (!q.empty()) q.pop();
}
void solve() {
  scanf("%d%d%d", &n, &m, &k);
  init();
  for (int i = 0, u, v; i < m; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v), g[v].push_back(u);
    deg[u]++, deg[v]++;
  }
  for (int i = 1; i <= n; i++)
    if (deg[i] < k) q.push(i), vis[i] = true;
  for (int i = 1; i <= n; i++) sort(g[i].begin(), g[i].end());
  while (!q.empty()) {
    int top = q.front();
    q.pop();
    now--;
    if (deg[top] == k - 1) {
      clique.clear();
      clique.push_back(top);
      for (int nxt : g[top])
        if (!del[nxt]) clique.push_back(nxt);
      int have_ans = 1;
      for (int i : clique) {
        for (int j : clique) {
          if (i == j) continue;
          if (!binary_search(g[i].begin(), g[i].end(), j)) {
            have_ans = 0;
            goto p1;
          }
        }
      }
    p1:
      if (have_ans) {
        printf("2\n");
        bool first = true;
        for (int i : clique) {
          if (first)
            printf("%d", i), first = false;
          else
            printf(" %d", i);
        }
        printf("\n");
        return;
      } else
        del[top] = 1;
    }
    del[top] = 1;
    for (int nxt : g[top]) {
      deg[nxt]--;
      if (deg[nxt] < k && !vis[nxt]) q.push(nxt), vis[nxt] = true;
    }
  }
  if (now) {
    printf("1 %d\n", now);
    for (int i = 1, cnt = 1; i <= n; i++, cnt++)
      if (!vis[i]) printf("%d%c", i, cnt == now ? '\n' : ' ');
  } else
    printf("-1\n");
}
int main() {
  cin >> T;
  while (T--) solve();
  return 0;
}