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 = 200500;
int n, m, k, T, hav[N], deg[N], ok, tot, cnt, pi[N];
priority_queue<pair<int, int> > q;
unordered_map<int, bool> ma[N];
vector<int> gg[N];
int head[N], pre[N << 1], to[N << 1];
inline void addedge(int u, int v) {
  pre[++cnt] = head[u];
  head[u] = cnt;
  to[cnt] = v;
}
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d%d", &n, &m, &k);
    while (!q.empty()) q.pop();
    for (int i = 1; i <= n; i++) ma[i].clear(), deg[i] = hav[i] = head[i] = 0;
    cnt = ok = tot = 0;
    for (int i = 1, u, v; i <= m; i++) {
      scanf("%d%d", &u, &v);
      ma[u][v] = ma[v][u] = 1;
      addedge(u, v);
      addedge(v, u);
      deg[u]++, deg[v]++;
    }
    for (int i = 1; i <= n; i++) q.push(make_pair(-deg[i], i));
    while (!q.empty()) {
      int now = q.top().second;
      q.pop();
      if (hav[now]) continue;
      hav[now] = 1;
      if (deg[now] < k - 1) {
        for (int i = head[now]; i; i = pre[i])
          if (!hav[to[i]]) deg[to[i]]--, q.push(make_pair(-deg[to[i]], to[i]));
      } else if (deg[now] > k - 1) {
        ok = cnt = 1;
        for (int i = 1; i <= n; i++)
          if (!hav[i]) cnt++;
        printf("1 %d\n%d ", cnt, now);
        for (int i = 1; i <= n; i++)
          if (!hav[i]) printf("%d ", i);
        printf("\n");
        break;
      } else {
        ++tot;
        gg[tot].clear();
        pi[tot] = now;
        for (int i = head[now]; i; i = pre[i])
          if (!hav[to[i]]) deg[to[i]]--, q.push(make_pair(-deg[to[i]], to[i]));
        for (int i = head[now]; i; i = pre[i])
          if (!hav[to[i]]) gg[tot].push_back(to[i]);
      }
    }
    if (ok) continue;
    for (int i = 1; i <= tot; i++) {
      int sz = gg[i].size(), f = 1;
      for (int j = 0; j < sz && f; j++)
        for (int k = j + 1; k < sz && f; k++)
          if (!ma[gg[i][j]][gg[i][k]]) f = 0;
      if (f) {
        ok = 1;
        printf("2\n%d ", pi[i]);
        for (int j = 0; j < sz; j++) printf("%d ", gg[i][j]);
        printf("\n");
        break;
      }
    }
    if (!ok) printf("-1\n");
  }
  return 0;
}