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 INF = 1e9;
struct SegTree {
  vector<int> minv, itree;
  int base;
  SegTree(int a, int arr[]) {
    base = 1;
    while (base < a) base <<= 1;
    minv.resize(base * 2, INF);
    itree.resize(base * 2);
    base--;
    for (int i = 1; i <= a; i++) {
      minv[base + i] = arr[i], itree[base + i] = i;
      if (!arr[i]) minv[base + i] = INF;
    }
    for (int i = base; i >= 1; i--) {
      minv[i] = min(minv[i * 2], minv[i * 2 + 1]);
      if (minv[i] == minv[i * 2])
        itree[i] = itree[i * 2];
      else
        itree[i] = itree[i * 2 + 1];
    }
  }
  void update(int idx, int val) {
    if (val == 0) val = INF;
    minv[idx += base] = val;
    idx >>= 1;
    while (idx) {
      minv[idx] = min(minv[idx * 2], minv[idx * 2 + 1]);
      if (minv[idx] == minv[idx * 2])
        itree[idx] = itree[idx * 2];
      else
        itree[idx] = itree[idx * 2 + 1];
      idx >>= 1;
    }
  }
  int get_idx() { return itree[1]; }
};
int tc, n, m, k;
unordered_set<int> lst[100005];
vector<int> ans;
int deg[100005];
bool in[100005];
void del_edge(int idx, SegTree &ST) {
  for (int v : lst[idx]) {
    if (deg[v]) {
      deg[v]--;
      ST.update(v, deg[v]);
    }
  }
  ST.update(idx, 0);
  deg[idx] = 0;
}
bool ctest(int v) {
  for (int x : ans) in[x] = false;
  ans.clear();
  ans.push_back(v);
  in[v] = true;
  for (int nxt : lst[v]) {
    if (deg[nxt]) ans.push_back(nxt), in[nxt] = true;
  }
  for (int x : ans) {
    int cnt = 0;
    for (int nxt : ans) {
      if (x != nxt && lst[x].count(nxt)) cnt++;
    }
    if (cnt != k - 1) return false;
  }
  return true;
}
void solve() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= n; i++) lst[i].clear(), deg[i] = 0;
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    lst[x].emplace(y);
    deg[x]++;
    lst[y].emplace(x);
    deg[y]++;
  }
  if (k * (k - 1) > 2 * m) {
    printf("-1\n");
    return;
  }
  SegTree ST(n, deg);
  for (int t = 1; t <= n; t++) {
    int idx = ST.get_idx();
    if (ST.minv[1] != INF) {
      if (deg[idx] < k - 1) {
        del_edge(idx, ST);
      } else if (deg[idx] == k - 1) {
        for (int i = 1; i <= n; i++) {
          if (deg[i] == k - 1) {
            bool tf = ctest(i);
            if (tf) {
              printf("2\n");
              for (int v : ans) printf("%d ", v);
              printf("\n");
              return;
            }
            del_edge(i, ST);
          }
        }
      } else {
        int cnt = 0;
        for (int i = 1; i <= n; i++)
          if (deg[i]) cnt++;
        printf("1 %d\n", cnt);
        for (int i = 1; i <= n; i++)
          if (deg[i]) printf("%d ", i);
        printf("\n");
        return;
      }
    } else
      break;
  }
  printf("-1\n");
}
int main() {
  for (scanf("%d", &tc); tc > 0; tc--) solve();
  return 0;
}