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;
int n, m, k, dg[100010], vis[100010], neig[100010], ecnt[100010], id[100010],
    a[100010], icnt;
queue<int> q;
vector<int> g[100010];
vector<pair<int, int> > vec[100010];
void check(int u, bool b) {
  for (int i = 0; i < g[u].size(); ++i) {
    int v = g[u][i];
    neig[v] = 1;
  }
  for (int i = 0; i < vec[u].size(); ++i) {
    if (neig[vec[u][i].first]) ecnt[vec[u][i].second]++;
  }
  for (int i = 0; i < g[u].size(); ++i) {
    int v = g[u][i];
    neig[v] = 0;
  }
  if (!b) return;
  ecnt[u] += k - 1;
  int acnt = 0;
  for (int i = 0; i < g[u].size(); ++i) {
    int v = g[u][i];
    if (!vis[v]) a[++acnt] = v;
  }
  for (int i = 1; i <= acnt; ++i) {
    int v = a[i];
    for (int j = i + 1; j <= acnt; ++j) {
      if (i == j) continue;
      int x = a[j];
      vec[v].push_back(make_pair(x, u));
    }
  }
}
int solve() {
  for (int i = 1; i <= n; ++i) {
    if (dg[i] < k) q.push(i);
  }
  int cnt = 0;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    check(u, dg[u] == k - 1);
    vis[u] = 1;
    id[u] = ++icnt;
    ++cnt;
    for (int i = 0; i < g[u].size(); ++i) {
      int v = g[u][i];
      if (vis[v]) continue;
      if (--dg[v] == k - 1) q.push(v);
    }
  }
  if (n - cnt >= k) {
    printf("1 %d\n", n - cnt);
    for (int i = 1; i <= n; ++i) {
      if (!vis[i]) printf("%d ", i);
    }
    puts("");
    return 1;
  }
  int mn = -1;
  for (int i = 1; i <= n; ++i) {
    if (ecnt[i] == (long long)k * (k - 1) / 2) {
      if (mn == -1 || id[mn] > id[i]) mn = i;
    }
  }
  if (mn != -1) {
    printf("2\n%d", mn);
    for (int j = 0; j < g[mn].size(); ++j) {
      if (id[g[mn][j]] >= id[mn]) printf(" %d", g[mn][j]);
    }
    puts("");
    return 2;
  }
  return -1;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    while (!q.empty()) q.pop();
    scanf("%d%d%d", &n, &m, &k);
    icnt = 0;
    for (int i = 1; i <= n; ++i) {
      vis[i] = ecnt[i] = neig[i] = dg[i] = 0;
      g[i].clear();
      vec[i].clear();
    }
    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);
      dg[u]++;
      dg[v]++;
    }
    if ((long long)k * (k - 1) / 2 > m) {
      puts("-1");
      continue;
    }
    int d = solve();
    if (d == -1) puts("-1");
  }
}