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 = 100005;
const int inf = 1000 * 1000 * 1000;
const int mod = 1000 * 1000 * 1000 + 7;
mt19937 myrand(chrono::steady_clock::now().time_since_epoch().count());
int t;
int n, m, k;
set<pair<int, int> > Q;
vector<int> g[N];
int sz[N];
bool used[N];
void clean(bool c) {
  while (!Q.empty()) {
    Q.erase(Q.begin());
  }
  for (int i = 1; i <= n; i++) {
    used[i] = false;
    if (c) {
      g[i].clear();
    }
  }
}
int main() {
  cin >> t;
  while (t--) {
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= m; i++) {
      int a, b;
      scanf("%d%d", &a, &b);
      g[a].push_back(b);
      g[b].push_back(a);
    }
    for (int i = 1; i <= n; i++) {
      sort(g[i].begin(), g[i].end());
      sz[i] = (int)g[i].size();
      Q.insert(make_pair((int)g[i].size(), i));
    }
    while (!Q.empty()) {
      int v = Q.begin()->second;
      if (sz[v] >= k) break;
      used[v] = true;
      Q.erase(Q.begin());
      for (int i = 0; i < (int)g[v].size(); i++) {
        int to = g[v][i];
        if (Q.find(make_pair(sz[to], to)) != Q.end()) {
          Q.erase(make_pair(sz[to], to));
          sz[to]--;
          Q.insert(make_pair(sz[to], to));
        }
      }
    }
    if ((int)Q.size() != 0) {
      vector<int> answ;
      for (int i = 1; i <= n; i++) {
        if (!used[i]) {
          answ.push_back(i);
        }
      }
      printf("1 %d\n", (int)answ.size());
      for (int i = 0; i < (int)answ.size(); i++) {
        printf("%d ", answ[i]);
      }
      printf("\n");
      clean(true);
      continue;
    }
    clean(false);
    if (k >= 500) {
      printf("-1\n");
      clean(true);
      continue;
    }
    for (int i = 1; i <= n; i++) {
      sz[i] = (int)g[i].size();
      Q.insert(make_pair((int)g[i].size(), i));
    }
    while (!Q.empty()) {
      int v = Q.begin()->second;
      if (sz[v] == k - 1) {
        vector<int> clique;
        for (int i = 0; i < (int)g[v].size(); i++) {
          if (Q.find(make_pair(sz[g[v][i]], g[v][i])) != Q.end()) {
            clique.push_back(g[v][i]);
          }
        }
        clique.push_back(v);
        bool flag = true;
        for (int i = 0; i < (int)clique.size(); i++) {
          int v = clique[i];
          for (int j = i + 1; j < (int)clique.size(); j++) {
            int ina = 0, inb = (int)g[v].size() - 1, res = -1;
            while (ina <= inb) {
              int mid = (ina + inb) / 2;
              if (g[v][mid] >= clique[j]) {
                res = mid;
                inb = mid - 1;
              } else {
                ina = mid + 1;
              }
            }
            if (res == -1 || g[v][res] != clique[j]) {
              flag = false;
              break;
            }
          }
          if (!flag) break;
        }
        if (flag) {
          printf("2\n");
          for (int i = 0; i < (int)clique.size(); i++) {
            printf("%d ", clique[i]);
          }
          printf("\n");
          break;
        }
      }
      Q.erase(Q.begin());
      for (int i = 0; i < (int)g[v].size(); i++) {
        int to = g[v][i];
        if (Q.find(make_pair(sz[to], to)) != Q.end()) {
          Q.erase(make_pair(sz[to], to));
          sz[to]--;
          Q.insert(make_pair(sz[to], to));
        }
      }
    }
    if ((int)Q.size() == 0) {
      printf("-1\n");
    }
    clean(true);
  }
  return 0;
}