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 = 200005;
int T, n, m, k, tot;
int deg[MAXN];
int a[MAXN];
bool vis[MAXN];
vector<int> e[MAXN];
unordered_map<int, bool> E[MAXN];
priority_queue<pair<int, int>, vector<pair<int, int> >,
               greater<pair<int, int> > >
    Q;
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= n; i++) {
      e[i].clear();
      E[i].clear();
      deg[i] = vis[i] = 0;
    }
    while (!Q.empty()) Q.pop();
    for (int u, v, i = 1; i <= m; i++) {
      scanf("%d%d", &u, &v);
      e[u].push_back(v);
      e[v].push_back(u);
      E[u][v] = E[v][u] = 1;
      deg[u]++;
      deg[v]++;
    }
    for (int i = 1; i <= n; i++) Q.push(make_pair(deg[i], i));
    bool gg = 0;
    while (!Q.empty()) {
      int u = Q.top().second;
      Q.pop();
      if (vis[u]) continue;
      bool killed = 0;
      if (deg[u] == k - 1) {
        bool ok = 1;
        if ((long long)k * (k - 1) / 2 <= m) {
          tot = 0;
          for (int i = 0; i < e[u].size(); i++) {
            int to = e[u][i];
            if (!vis[to]) a[++tot] = to;
          }
          for (int i = 1; ok && i <= tot; i++)
            for (int j = i + 1; ok && j <= tot; j++)
              if (!E[a[i]][a[j]]) ok = 0;
        } else
          ok = 0;
        if (ok) {
          printf("2\n%d ", u);
          for (int i = 0; i < e[u].size(); i++) {
            int to = e[u][i];
            if (!vis[to]) printf("%d ", to);
          }
          printf("\n");
          gg = 1;
          break;
        }
        killed = 1;
      }
      if (deg[u] < k - 1 || killed) {
        for (int i = 0; i < e[u].size(); i++) {
          int to = e[u][i];
          if (!vis[to]) {
            deg[to]--;
            Q.push(make_pair(deg[to], to));
          }
        }
      }
      if (deg[u] >= k) {
        int tot = 0;
        for (int i = 1; i <= n; i++)
          if (!vis[i]) tot++;
        printf("1 %d\n", tot);
        for (int i = 1; i <= n; i++)
          if (!vis[i]) printf("%d ", i);
        printf("\n");
        gg = 1;
        break;
      }
      vis[u] = 1;
    }
    if (!gg) puts("-1");
  }
  return 0;
}