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;
inline long long read() {
  long long x = 0;
  char ch = getchar();
  bool d = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') d = 0;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return d ? x : -x;
}
inline unsigned long long rnd() {
  return ((unsigned long long)rand() << 30 ^ rand()) << 4 | rand() % 4;
}
const int N = 1e5 + 5;
vector<long long> v[N];
int du[N];
bool used[N];
queue<int> q;
bool check(int u) {
  vector<long long> rv(1);
  rv[0] = u;
  for (auto x : v[u])
    if (!used[x]) rv.push_back(x);
  for (auto x : rv) {
    for (auto y : rv)
      if (x != y) {
        int pos = lower_bound(v[x].begin(), v[x].end(), y) - v[x].begin();
        if (pos == (int)v[x].size() || v[x][pos] != y) return 0;
      }
  }
  return 1;
}
void solve(int n, int m, int k) {
  for (int i = (int)(1); i <= (int)(n); i++) v[i].clear(), du[i] = used[i] = 0;
  while (q.size()) q.pop();
  vector<long long> ans;
  for (int i = (int)(1); i <= (int)(m); i++) {
    int x = read(), y = read();
    v[x].push_back(y);
    v[y].push_back(x);
    du[x]++;
    du[y]++;
  }
  int cnt = 0;
  for (int i = (int)(1); i <= (int)(n); i++)
    if (du[i] < k) cnt++, used[i] = 1, q.push(i);
  while (q.size()) {
    int u = q.front();
    q.pop();
    for (auto x : v[u]) {
      du[x]--;
      if (!used[x] && du[x] < k) cnt++, used[x] = 1, q.push(x);
    }
  }
  if (cnt != n) {
    cout << 1 << " " << n - cnt << endl;
    for (int i = (int)(1); i <= (int)(n); i++)
      if (!used[i]) printf("%d ", i);
    puts("");
    return;
  }
  for (int i = (int)(1); i <= (int)(n); i++) sort(v[i].begin(), v[i].end());
  for (int i = (int)(1); i <= (int)(n); i++)
    used[i] = 0, du[i] = (int)v[i].size();
  for (int i = (int)(1); i <= (int)(n); i++)
    if (du[i] < k - 1) q.push(i), used[i] = 1;
  while (q.size()) {
    int u = q.front();
    q.pop();
    for (auto x : v[u]) {
      du[x]--;
      if (!used[x] && du[x] < k - 1) used[x] = 1, q.push(x);
    }
  }
  for (int i = (int)(1); i <= (int)(n); i++)
    if (du[i] == k - 1) q.push(i);
  while (q.size()) {
    int u = q.front();
    q.pop();
    if (du[u] == k - 1 && check(u)) {
      puts("2");
      printf("%d ", u);
      for (auto x : v[u])
        if (!used[x]) printf("%d ", x);
      puts("");
      return;
    }
    used[u] = 1;
    for (auto x : v[u]) {
      du[x]--;
      if (!used[x] && du[x] == k - 1) q.push(x);
    }
  }
  puts("-1");
  return;
}
int main() {
  int T = read();
  while (T--) {
    int n = read(), m = read(), k = read();
    solve(n, m, k);
  }
}