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 = 100050;
const int inf = 0x3f3f3f3f;
int n, m;
vector<int> e[maxn];
int du[maxn];
int inqueue[maxn];
void init(int n) {
  for (int i = 0; i <= n; i++) {
    e[i].clear();
    inqueue[i] = 0;
    du[i] = 0;
  }
}
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  long long f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f *= -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x *= f;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tn;
  read(tn);
  while (tn--) {
    int k;
    read(n), read(m), read(k);
    init(n);
    for (int i = 1; i <= m; i++) {
      int u, v;
      read(u), read(v);
      e[u].push_back(v), e[v].push_back(u);
      du[u]++, du[v]++;
    }
    for (int i = 1; i <= n; i++) {
      sort(e[i].begin(), e[i].end());
    }
    queue<int> q;
    for (int i = 1; i <= n; i++) {
      if (du[i] < k) q.push(i), inqueue[i] = 1;
    }
    vector<int> c, ans;
    while (q.size()) {
      int u = q.front();
      q.pop();
      inqueue[u] = 2;
      for (auto i : e[u]) {
        du[i]--;
        if (inqueue[i]) continue;
        if (du[i] < k) q.push(i), inqueue[i] = 1;
      }
      if (m * 2 >= k * (k - 1)) {
        if (du[u] == k - 1 && c.empty()) {
          c.push_back(u);
          for (auto i : e[u])
            if (inqueue[i] != 2) c.push_back(i);
          bool ok = 1;
          for (int i = 0; i < c.size(); i++) {
            for (int j = i + 1; j < c.size(); j++) {
              if (!binary_search(e[c[i]].begin(), e[c[i]].end(), c[j])) {
                ok = 0;
                break;
              }
            }
            if (!ok) break;
          }
          if (!ok) c.clear();
        }
      }
    }
    for (int i = 1; i <= n; i++) {
      if (!inqueue[i]) ans.push_back(i);
    }
    if (ans.size() >= k) {
      printf("1 %d\n", ans.size());
      for (int i = 0; i < ans.size(); i++) {
        if (i) putchar(' ');
        printf("%d", ans[i]);
      }
      putchar('\n');
    } else if (c.size()) {
      printf("2\n");
      for (int i = 0; i < c.size(); i++) {
        if (i) putchar(' ');
        printf("%d", c[i]);
      }
      putchar('\n');
    } else
      printf("-1\n");
  }
  return 0;
}