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 N = 111111;
int n, m, k, a[N], b[N], fl;
vector<int> v[N], ans;
long long p[N];
bool chk(int x, int y) {
  if (x > y) swap(x, y);
  long long z = (long long)x << 32 | y;
  return (*lower_bound(p, p + m, z)) == z;
}
void del(int x) {
  if (!b[x] || fl) return;
  b[x] = 0;
  int i, j, y, f;
  vector<int> t;
  for (i = 0; i < v[x].size(); i = i + 1) {
    y = v[x][i];
    if (b[y]) {
      t.push_back(y);
      a[y]--;
    }
  }
  if (t.size() == k - 1 && k <= 500 && k * (k - 1) / 2 <= m) {
    f = 1;
    for (i = 0; f && i < k - 1; i = i + 1)
      for (j = i + 1; f && j < k - 1; j = j + 1)
        if (!chk(t[i], t[j])) f = 0;
    if (f && !fl) {
      fl = 1;
      ans = t;
      ans.push_back(x);
    }
  }
  for (i = 0; i < t.size(); i = i + 1) {
    y = t[i];
    if (a[y] < k) del(y);
  }
}
int main() {
  int T, i, x, y;
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d%d", &n, &m, &k);
    for (i = 1; i <= n; i = i + 1) {
      a[i] = 0, b[i] = 1;
      v[i].clear();
    }
    fl = 0, ans.clear();
    for (i = 0; i < m; i = i + 1) {
      scanf("%d%d", &x, &y);
      if (x > y) swap(x, y);
      p[i] = (long long)x << 32 | y;
      a[x]++, a[y]++;
      v[x].push_back(y);
      v[y].push_back(x);
    }
    sort(p, p + m), p[m] = (long long)N << 32;
    for (i = 1; i <= n; i = i + 1)
      if (a[i] < k) del(i);
    if (fl) {
      printf("2\n");
      for (i = 0; i < ans.size(); i = i + 1) printf("%d ", ans[i]);
      printf("\n");
      continue;
    }
    x = 0;
    for (i = 1; i <= n; i = i + 1)
      if (b[i]) x++, ans.push_back(i);
    if (x) {
      printf("1 %d\n", ans.size());
      for (i = 0; i < ans.size(); i = i + 1) printf("%d ", ans[i]);
      printf("\n");
      continue;
    }
    printf("-1\n");
  }
  return 0;
}