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 = 100003;
vector<int> g[maxn];
int n, m, k, vis[maxn], deg[maxn], p[maxn];
bool ok;
void chk(int v) {
  if (1ll * k * (k - 1) / 2 <= m) {
    bool flag = 1;
    int cnt = 0;
    p[++cnt] = v;
    for (int w : g[v])
      if (vis[w] != 2) p[++cnt] = w;
    for (int w : g[v])
      if (vis[w] != 2) {
        for (int j = 1; j <= k; j++)
          if (p[j] != w) {
            if (!binary_search(g[w].begin(), g[w].end(), p[j])) {
              flag = 0;
              break;
            }
          }
        if (!flag) break;
      }
    if (flag) {
      ok = 1;
      puts("2");
      for (int j = 1; j <= k; j++) printf("%d ", p[j]);
      puts("");
    }
  }
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= m; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      g[u].push_back(v), g[v].push_back(u);
    }
    ok = 0;
    queue<int> q;
    for (int i = 1; i <= n; i++) {
      sort(g[i].begin(), g[i].end());
      deg[i] = g[i].size();
      if (deg[i] < k) vis[i] = 1, q.push(i);
    }
    if (!ok) {
      while (!q.empty()) {
        int u = q.front(), cnt = g[u].size();
        q.pop();
        vis[u] = 2;
        for (int v : g[u]) cnt -= (vis[v] == 2);
        if (cnt == k - 1) chk(u);
        for (int v : g[u])
          if (!vis[v])
            if (--deg[v] < k) vis[v] = 1, q.push(v);
        if (ok) break;
      }
      if (!ok) {
        for (int i = 1; i <= n; i++)
          if (!vis[i] && deg[i] < k) vis[i] = 1, q.push(i);
        while (!q.empty()) {
          int u = q.front();
          q.pop();
          for (int v : g[u])
            if (!vis[v])
              if (--deg[v] < k) vis[v] = 1, q.push(v);
        }
        int cnt = 0;
        for (int i = 1; i <= n; i++) cnt += !vis[i];
        if (cnt) {
          printf("1 %d\n", cnt);
          for (int i = 1; i <= n; i++)
            if (!vis[i]) printf("%d ", i);
          puts("");
        } else
          puts("-1");
      }
    }
    for (int i = 1; i <= n; i++) g[i].clear(), vis[i] = 0;
  }
  return 0;
}