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;
inline int read() {
  int x = 0;
  bool flg = false;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') flg = true;
  for (; isdigit(ch); ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ 48);
  return flg ? -x : x;
}
const int mod = 1e6 + 3;
int n, m, k;
vector<int> e[100010];
int d[100010];
set<pair<int, int> > st[1 << 20];
vector<int> work(int k) {
  static int d[100010], que[100010], hd, tl;
  static int vis[100010], tim;
  for (int i = 1; i <= n; i++) d[i] = ::d[i];
  hd = tl = 0;
  ++tim;
  for (int i = 1; i <= n; i++)
    if (d[i] < k) vis[i] = tim, que[++tl] = i;
  while (hd < tl) {
    int u = que[++hd];
    for (int x : e[u])
      if (vis[x] != tim && (--d[x]) < k) que[++tl] = x, vis[x] = tim;
  }
  vector<int> S;
  for (int i = 1; i <= n; i++)
    if (vis[i] ^ tim) S.push_back(i);
  return S;
}
int c[100010];
bool check(int x) {
  vector<int> S;
  S.reserve(k);
  S.push_back(x);
  for (int y : e[x])
    if (c[y] >= 1) S.push_back(y);
  if (S.size() < k) return 0;
  if (S.size() != k) printf("%d :: %d\n", k, int(S.size())), exit(1234);
  for (int x : S)
    for (int y : S)
      if (x < y) {
        int h = (1LL * x * x * x + 1LL * y * y * y) % mod;
        if (!st[h].count((pair<int, int>){x, y})) return 0;
      }
  return 1;
}
void init() {
  for (int i = 1; i <= n; i++) {
    for (int x : e[i])
      if (x < i) {
        int h = (1LL * i * i * i + 1LL * x * x * x) % mod;
        st[h].clear();
      }
    e[i].clear(), d[i] = 0;
  }
}
void solve() {
  n = read(), m = read(), k = read();
  for (int i = 1; i <= m; i++) {
    int u = read(), v = read();
    if (u > v) swap(u, v);
    e[u].push_back(v);
    e[v].push_back(u);
    d[u]++, d[v]++;
    int h = (1LL * u * u * u + 1LL * v * v * v) % mod;
    st[h].insert((pair<int, int>){u, v});
  }
  auto S = work(k);
  if (!S.empty()) {
    printf("1 %d\n", int(S.size()));
    for (int x : S) printf("%d ", x);
    puts("");
    return;
  }
  S = work(k - 1);
  if (S.empty()) return puts("-1"), void();
  static int que[100010], hd, tl;
  hd = tl = 0;
  for (int i = 1; i <= n; i++) c[i] = 0, d[i] = 0;
  for (int x : S) c[x] = 1;
  for (int x : S) {
    d[x] = 0;
    for (int y : e[x])
      if (c[y] >= 1) d[x]++;
    if (d[x] < k - 1) exit(2345);
    if (d[x] == k - 1) que[++tl] = x, c[x] = 2;
  }
  while (hd < tl) {
    int u = que[++hd];
    if (c[u] && check(u)) {
      puts("2");
      printf("%d", u);
      for (int x : e[u])
        if (c[x] >= 1) printf(" %d", x);
      puts("");
      return;
    }
    c[u] = 0;
    for (int x : e[u]) {
      if (!c[x]) continue;
      --d[x];
      if (d[x] < k - 1) {
        if (c[x] != 2) exit(3456);
        c[x] = 0;
      } else if (d[x] < k) {
        if (c[x] != 1) exit(4567);
        c[x] = 2, que[++tl] = x;
      }
    }
  }
  puts("-1");
}
int main() {
  for (int T = read(); T--;) init(), solve();
  return 0;
}