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>
const int maxn = 2e5 + 10;
const int mod = 998244353;
using namespace std;
template <typename T1, typename T2>
void ckmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
void ckmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
template <typename T1, typename T2>
void addmo(T1 &a, T2 b) {
  a += b;
  if (a >= mod) a -= mod;
}
int n, m, k, vis[maxn], deg[maxn], del[maxn], ty;
vector<int> g[maxn], t1, t2;
int bfs() {
  queue<int> q;
  for (int i = 1; i <= n; i++) {
    if (deg[i] < k) {
      vis[i] = 1;
      q.push(i);
    }
  }
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    if (deg[x] == k - 1) {
      t1.clear();
      t1.push_back(x);
      for (int i : g[x])
        if (!del[i]) {
          t1.push_back(i);
        }
      bool ok = true;
      for (int i : t1)
        for (int j : t1) {
          if (i == j) break;
          auto it = lower_bound(g[i].begin(), g[i].end(), j);
          if (it == g[i].end() || *it != j) {
            ok = false;
          }
        }
      if (ok) return ty = 1;
    }
    del[x] = 1;
    for (int i : g[x]) {
      deg[i]--;
      if (deg[i] < k && !vis[i]) {
        vis[i] = 1;
        q.push(i);
      }
    }
  }
  t2.clear();
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) t2.push_back(i), ty = 2;
  }
  return ty;
}
void solve() {
  ty = 0;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    g[i].clear();
    deg[i] = 0, vis[i] = 0, del[i] = 0;
  }
  for (int i = 1, x, y; i <= m; i++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y), g[y].push_back(x);
    deg[x]++, deg[y]++;
  }
  if (1LL * k * (k - 1) / 2 > m) return printf("-1\n"), void();
  for (int i = 1; i <= n; i++) sort(g[i].begin(), g[i].end());
  if (!bfs()) return printf("-1\n"), void();
  if (ty == 1) {
    printf("2\n");
    for (int i : t1) printf("%d ", i);
  } else {
    printf("1 %d\n", t2.size());
    for (int i : t2) printf("%d ", i);
  }
  printf("\n");
}
signed main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}