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 + 10;
const int INF = 2e9;
const double PI = acos(-1);
const double eps = 1e-6;
int T, n, m, k;
int x[MAXN], y[MAXN], ind[MAXN];
bool del[MAXN], can[MAXN];
vector<int> nodes;
unordered_set<int> G[MAXN];
set<pair<int, int> > S;
template <typename T>
inline bool read(T &a) {
  a = 0;
  char c = getchar();
  int f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    a = a * 10 + (c ^ 48);
    c = getchar();
  }
  a *= f;
  return 1;
}
template <typename A, typename... B>
inline bool read(A &x, B &...y) {
  return read(x) && read(y...);
}
void solve() {
  read(n), read(m), read(k);
  for (int i = 1; i <= n; ++i) G[i].clear(), ind[i] = 0;
  S.clear();
  nodes.clear();
  for (int i = 1; i <= m; ++i) {
    read(x[i], y[i]), ind[x[i]]++, ind[y[i]]++;
    G[x[i]].insert(y[i]);
    G[y[i]].insert(x[i]);
  }
  if (k > 500) return puts("-1"), void();
  for (int i = 1; i <= n; ++i)
    S.insert(make_pair(ind[i], i)), del[i] = can[i] = false;
  while (S.size()) {
    auto now = *S.begin();
    S.erase(S.begin());
    if (now.first == k - 1) {
      int x = now.second;
      bool flag = true;
      for (auto a : G[x]) {
        for (auto b : G[x]) {
          if (a == b) continue;
          if (!G[a].count(b)) {
            flag = false;
            break;
          }
        }
        if (!flag) break;
      }
      if (flag) {
        printf("2\n%d ", x);
        for (auto a : G[x]) printf("%d ", a);
        printf("\n");
        return;
      }
    }
    if (now.first <= k - 1) {
      del[now.second] = true;
      for (auto to : G[now.second]) {
        if (del[to]) continue;
        S.erase(S.find(make_pair(ind[to], to)));
        ind[to]--;
        G[to].erase(now.second);
        if (!ind[to]) continue;
        S.insert(make_pair(ind[to], to));
      }
      G[now.second].clear();
    } else {
      printf("1 %d\n", S.size() + 1);
      printf("%d ", now.second);
      for (auto x : S) printf("%d ", x.second);
      printf("\n");
      return void();
    }
  }
  puts("-1");
  return void();
}
signed main() {
  read(T);
  while (T--) solve();
  return 0;
}