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 = 2e5 + 8, inf = 0x3f3f3f3f, mod = 1e9 + 7;
template <typename T1, typename T2>
struct Table {
  T1 key[maxn];
  T2 val[maxn];
  int head[19260817], next[maxn], tot;
  int sta[maxn], top;
  void init(bool flag) {
    if (flag) {
      memset(head, -1, sizeof(head));
      tot = top = 0;
      return;
    }
    while (top > 0) head[sta[--top]] = -1;
    tot = 0;
  }
  int gethas(T1 v) { return (v.first * 1000 + v.second) % 19260817; }
  void modify(T1 x, T2 y) {
    int u = gethas(x);
    for (int i = head[u]; ~i; i = next[i]) {
      if (key[i] == x) {
        val[i] = y;
        return;
      }
    }
    key[tot] = x;
    val[tot] = y;
    next[tot] = head[u];
    if (head[u] == -1) {
      sta[top++] = u;
    }
    head[u] = tot++;
  }
  T2 query(T1 x) {
    int u = gethas(x);
    for (int i = head[u]; ~i; i = next[i]) {
      if (key[i] == x) {
        return val[i];
      }
    }
    return 0;
  }
};
Table<pair<int, int>, int> es;
vector<int> g[maxn];
struct node {
  int d, id;
  bool operator<(const node &rhs) const { return d > rhs.d; }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int kase;
  cin >> kase;
  es.init(1);
  while (kase--) {
    es.init(0);
    int n, m, k;
    cin >> n >> m >> k;
    for (int i = 1; i <= n; ++i) g[i].clear();
    for (int i = 0, u, v; i < m; ++i) {
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
      es.modify(pair<int, int>(u, v), 1);
      es.modify(pair<int, int>(v, u), 1);
    }
    vector<int> deg(n + 1), vis(n + 1);
    priority_queue<node> pq;
    for (int i = 1; i <= n; ++i) {
      sort(g[i].begin(), g[i].end());
      deg[i] = g[i].size();
      if (deg[i] < k) pq.push({deg[i], i}), vis[i] = 1;
    }
    bool ok = 0;
    while (!pq.empty()) {
      auto cur = pq.top();
      pq.pop();
      int d = cur.d, u = cur.id;
      vis[u] = -1;
      if (d == k - 1 && 1ll * k * (k - 1) / 2 <= m) {
        vector<int> clique;
        clique.push_back(u);
        for (auto &v : g[u])
          if (vis[v] != -1) clique.push_back(v);
        if (clique.size() == k) {
          bool is = 1;
          for (auto &i1 : clique) {
            for (auto &i2 : clique) {
              if (i1 == i2) continue;
              if (!es.query(pair<int, int>(i1, i2))) {
                is = 0;
                break;
              }
            }
            if (!is) break;
          }
          if (is) {
            ok = 1;
            cout << "2\n";
            for (auto &v : clique) cout << v << ' ';
            cout << '\n';
            break;
          }
        }
      }
      for (auto &v : g[u]) {
        if (vis[v] == 0 && --deg[v] < k) {
          pq.push({deg[v], v});
          vis[v] = 1;
        }
      }
    }
    if (!ok) {
      bool can = 0;
      vector<int> ans;
      for (int i = 1; i <= n; ++i)
        if (vis[i] == 0) can = 1, ans.push_back(i);
      if (can) {
        cout << "1 " << ans.size() << '\n';
        for (auto &v : ans) cout << v << ' ';
        cout << '\n';
      } else
        cout << "-1\n";
    }
  }
  return 0;
}