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 + 5;
const int inf = 0x3f3f3f3f;
const int modu = 1e9 + 7;
struct Node {
  int d, x;
  Node(int d = 0, int x = 0) : d(d), x(x) {}
  bool operator<(const Node &op) const { return d > op.d; }
};
int n, m, k;
int deg[maxn];
vector<int> g[maxn], res;
unordered_set<int> f[maxn];
priority_queue<Node> Q;
bool vis[maxn];
bool check(int x) {
  res.clear();
  for (int i = 0; i < g[x].size(); ++i)
    if (!vis[g[x][i]]) res.push_back(g[x][i]);
  for (int i = 0; i < res.size(); ++i)
    for (int j = i + 1; j < res.size(); ++j)
      if (!f[res[i]].count(res[j])) return false;
  res.push_back(x);
  return true;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= n; ++i) {
      g[i].clear();
      f[i].clear();
      deg[i] = 0;
      vis[i] = false;
    }
    for (int i = 0; i < m; ++i) {
      int u, v;
      scanf("%d%d", &u, &v);
      g[u].push_back(v);
      f[u].insert(v);
      g[v].push_back(u);
      f[v].insert(u);
      deg[u]++;
      deg[v]++;
    }
    if (k >= sqrt(2 * m) + 1) {
      printf("-1\n");
      continue;
    }
    for (int i = 1; i <= n; ++i) sort(g[i].begin(), g[i].end());
    while (!Q.empty()) Q.pop();
    for (int i = 1; i <= n; ++i) Q.push(Node(deg[i], i));
    int ans = -1, st = 0;
    while (!Q.empty()) {
      int x = Q.top().x;
      Q.pop();
      while (!Q.empty() && vis[x]) {
        x = Q.top().x;
        Q.pop();
        if (Q.empty()) break;
      }
      if (vis[x]) break;
      if (deg[x] >= k) {
        ans = 1;
        st = 0;
        for (int i = 1; i <= n; ++i)
          if (!vis[i]) st++;
        break;
      } else if (deg[x] == k - 1 && check(x)) {
        ans = 2;
        break;
      }
      vis[x] = true;
      for (int i = 0; i < g[x].size(); ++i) {
        int y = g[x][i];
        if (!vis[y]) {
          deg[y]--;
          Q.push(Node(deg[y], y));
        }
      }
    }
    if (ans == 1) {
      printf("1 %d\n", st);
      for (int i = 1; i <= n; ++i)
        if (!vis[i]) printf("%d ", i);
    } else if (ans == 2) {
      printf("2\n");
      for (int i = 0; i < res.size(); ++i) printf("%d ", res[i]);
    } else
      printf("-1");
    printf("\n");
  }
  return 0;
}