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 + 10;
char buf[100000], *p1 = buf, *p2 = buf;
inline char nc() {
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int read() {
  int w = 1, data = 0;
  char ch = 0;
  while (ch != '-' && (ch < '0' || ch > '9')) ch = nc();
  if (ch == '-') w = -1, ch = nc();
  while (ch >= '0' && ch <= '9') data = data * 10 + ch - '0', ch = nc();
  return w * data;
}
int n, m, k;
vector<int> du(MAXN), vis(MAXN);
vector<vector<int>> g(MAXN);
void init() {
  for (int i = 0; i <= n; i++) g[i].clear(), du[i] = 0, vis[i] = 0;
}
pair<int, vector<int>> check() {
  if (k * (k - 1) > 2 * m) return make_pair(0, vector<int>{});
  queue<int> q;
  for (int i = 1; i <= n; i++)
    if (du[i] < k) q.push(i);
  while (!q.empty()) {
    int t = q.front();
    q.pop();
    if (vis[t]) continue;
    vis[t] = 1;
    if (du[t] == k - 1) {
      vector<int> a{t};
      for (auto x : g[t])
        if (!vis[x]) a.push_back(x);
      bool flag = 1;
      for (int i = 0; i < (int)a.size(); i++) {
        for (int j = i + 1; j < (int)a.size(); j++) {
          if (binary_search(g[a[i]].begin(), g[a[i]].end(), a[j]) == 0) {
            flag = 0;
            break;
          }
        }
      }
      if (flag == 1) return make_pair(2, a);
    }
    for (auto v : g[t]) {
      du[v]--;
      if (vis[v]) continue;
      if (du[v] < k) q.push(v);
    }
  }
  vector<int> a;
  for (int i = 1; i <= n; i++)
    if (vis[i] == 0) a.push_back(i);
  if (a.size()) return make_pair(1, a);
  return make_pair(0, vector<int>{});
}
void doit() {
  init();
  n = read();
  m = read();
  k = read();
  for (int i = 1; i <= m; i++) {
    int x = read(), y = read();
    g[x].push_back(y);
    g[y].push_back(x);
    du[x]++;
    du[y]++;
  }
  for (int i = 1; i <= n; i++) sort(g[i].begin(), g[i].end());
  pair<int, vector<int>> res = check();
  if (res.first == 0)
    printf("-1\n");
  else if (res.first == 1) {
    printf("1 %d\n", (int)res.second.size());
    for (auto x : res.second) printf("%d ", x);
    printf("\n");
  } else if (res.first == 2) {
    printf("2\n");
    for (auto x : res.second) printf("%d ", x);
    printf("\n");
  }
}
int main() {
  int _ = read();
  while (_--) doit();
  return 0;
}