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;
int read() {
  int s = 0;
  char c = getchar(), lc = '+';
  while (c < '0' || '9' < c) lc = c, c = getchar();
  while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
  return lc == '-' ? -s : s;
}
void write(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x < 10)
    putchar(x + '0');
  else {
    write(x / 10);
    putchar(x % 10 + '0');
  }
}
void print(int x, char c = '\n') {
  write(x);
  putchar(c);
}
vector<int> e[N];
int siz[N], q[N], k;
int vis[N], p[N], cnt;
bool check(int now) {
  int cnt = 0;
  for (int i = 0; i < e[now].size(); i++) {
    int to = e[now][i];
    if (vis[to] == 2) continue;
    if (siz[to] < k - 1) return 0;
    p[++cnt] = to;
  }
  for (int i = 1; i < cnt; i++)
    for (int j = i + 1; j <= cnt; j++)
      if (!binary_search(e[p[i]].begin(), e[p[i]].end(), p[j])) return 0;
  print(2);
  print(now, ' ');
  for (int i = 1; i <= cnt; i++) print(p[i], ' ');
  puts("");
  return 1;
}
signed main() {
  int T = read();
  while (T--) {
    int n = read(), m = read(), h = 1, t = 0;
    k = read();
    for (int i = 1; i <= n; i++) e[i].clear();
    for (int i = 1; i <= n; i++) siz[i] = 0;
    for (int i = 1; i <= n; i++) vis[i] = 0;
    for (int i = 1; i <= m; i++) {
      int u = read(), v = read();
      e[u].push_back(v);
      e[v].push_back(u);
      siz[u]++;
      siz[v]++;
    }
    if (k * (k - 1) > 2 * m) {
      puts("-1");
      continue;
    }
    for (int i = 1; i <= n; i++) sort(e[i].begin(), e[i].end());
    bool flag = 0;
    for (int i = 1; i <= n; i++)
      if (siz[i] < k) {
        q[++t] = i;
        vis[i] = 1;
      }
    if (flag) continue;
    while (h <= t) {
      int now = q[h++];
      vis[now] = 2;
      if (siz[now] == k - 1)
        if (check(now)) {
          flag = 1;
          break;
        }
      for (int i = 0; i < e[now].size(); i++) {
        int to = e[now][i];
        siz[now]--;
        siz[to]--;
        if (siz[to] < k && !vis[to]) {
          vis[to] = 1;
          q[++t] = to;
        }
      }
    }
    if (flag) continue;
    int cnt = 0;
    for (int i = 1; i <= n; i++)
      if (!vis[i]) cnt++;
    if (cnt) {
      print(1, ' ');
      print(cnt);
      for (int i = 1; i <= n; i++)
        if (!vis[i]) print(i, ' ');
      puts("");
    } else
      puts("-1");
  }
  return 0;
}