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 = 1e5 + 5;
vector<int> edge[N];
void add(int u, int v) { edge[u].push_back(v); }
bool find(int u, int v) {
  vector<int>::iterator it;
  it = lower_bound(edge[u].begin(), edge[u].end(), v);
  return it != edge[u].end() && *it == v;
}
int in[N];
int a[N];
bool vis[N];
void solve() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  bool flag = (long long)k * (k - 1) / 2 <= m;
  for (int i = 1; i <= n; ++i) {
    in[i] = 0;
    vis[i] = 0;
    vector<int>().swap(edge[i]);
  }
  for (int i = 1; i <= m; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
    ++in[u], ++in[v];
  }
  for (int i = 1; i <= n; ++i) {
    sort(edge[i].begin(), edge[i].end());
  }
  queue<int> q;
  for (int i = 1; i <= n; ++i) {
    if (in[i] < k) {
      q.push(i);
    }
  }
  while (!q.empty()) {
    int t = q.front();
    q.pop();
    if (vis[t]) {
      continue;
    }
    if (in[t] == k - 1 && flag) {
      int tot = 0;
      for (int i = 0; i < (int)edge[t].size(); ++i) {
        int to = edge[t][i];
        if (vis[to]) {
          continue;
        }
        a[++tot] = to;
      }
      bool is = 0;
      if (tot != k - 1) {
        is = 1;
      }
      for (int i = 1; i <= tot && !is; ++i) {
        for (int j = i + 1; j <= tot; ++j) {
          if (!find(a[i], a[j])) {
            is = 1;
            break;
          }
        }
      }
      if (!is) {
        printf("2\n");
        for (int i = 1; i <= tot; ++i) {
          printf("%d ", a[i]);
        }
        printf("%d\n", t);
        return;
      }
    }
    vis[t] = 1;
    for (int i = 0; i < (int)edge[t].size(); ++i) {
      int to = edge[t][i];
      if (vis[to]) {
        continue;
      }
      --in[to];
      if (in[to] < k) {
        q.push(to);
      }
    }
  }
  int t = 0;
  for (int i = 1; i <= n; ++i) {
    if (!vis[i]) {
      a[++t] = i;
    }
  }
  if (t) {
    printf("1 %d\n", t);
    for (int i = 1; i <= t; ++i) {
      printf("%d ", a[i]);
    }
    printf("\n");
  } else {
    printf("-1\n");
  }
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    solve();
  }
}