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;
int T, n, m, k;
int last[200007], pre[200007], other[200007], cnt = 0, d[200007];
int sz;
set<pair<int, int> > s;
unordered_map<long long, bool> mp;
bool vis[200007];
int que[200007], tot;
void insert(int u, int v) {
  other[++cnt] = v, pre[cnt] = last[u], last[u] = cnt;
}
bool check(int u, int v) {
  if (mp.find(1LL * n * u + v) != mp.end()) return true;
  if (mp.find(1LL * n * v + u) != mp.end()) return true;
  return false;
}
void solve() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) last[i] = 0, d[i] = 0, vis[i] = false;
  mp.clear();
  s.clear();
  cnt = 0;
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    mp[1LL * n * u + v] = true;
    mp[1LL * n * v + u] = true;
    insert(u, v), insert(v, u);
    ++d[u], ++d[v];
  }
  for (int i = 1; i <= n; i++) s.insert(make_pair(d[i], i));
  sz = n;
  if ((1LL * k * (k - 1)) / 2 > m) {
    printf("%d\n", -1);
    return;
  }
  while (sz > 0 && (*s.begin()).first <= k - 1) {
    if ((*s.begin()).first < k - 1) {
      int u = (*s.begin()).second;
      s.erase(s.begin());
      vis[u] = true;
      --sz;
      for (int q = last[u]; q; q = pre[q]) {
        int v = other[q];
        if (vis[v]) continue;
        --d[u];
        --d[v];
        s.erase(make_pair(d[v] + 1, v));
        s.insert(make_pair(d[v], v));
      }
    } else {
      int u = (*s.begin()).second;
      tot = 1;
      que[1] = u;
      for (int q = last[u]; q; q = pre[q]) {
        int v = other[q];
        if (!vis[v]) que[++tot] = v;
      }
      random_shuffle(que + 1, que + tot + 1);
      bool flag = true;
      for (int i = 1; i <= tot && flag; i++)
        for (int j = i + 1; j <= tot && flag; j++)
          if (!check(que[i], que[j])) {
            flag = false;
            break;
          }
      if (flag) {
        printf("%d\n", 2);
        for (int i = 1; i <= tot; i++) printf("%d ", que[i]);
        printf("\n");
        return;
      } else {
        vis[u] = true;
        s.erase(s.begin());
        --sz;
        for (int q = last[u]; q; q = pre[q]) {
          int v = other[q];
          if (vis[v]) continue;
          --d[u];
          --d[v];
          s.erase(make_pair(d[v] + 1, v));
          s.insert(make_pair(d[v], v));
        }
      }
    }
  }
  if (sz > 0) {
    printf("%d %d\n", 1, sz);
    for (set<pair<int, int> >::iterator it = s.begin(); it != s.end(); it++)
      printf("%d ", (*it).second);
    printf("\n");
  } else
    printf("%d\n", -1);
}
int main() {
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}