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 long long gcd(long long a, long long b) {
  while (b != 0) {
    long long c = a % b;
    a = b;
    b = c;
  }
  return a < 0 ? -a : a;
}
inline long long lowbit(long long x) { return x & (-x); }
int head[2000010], Edge_Num;
struct Edge {
  int to, next;
  long long w;
} e[4000010];
inline void ade(int x, int y, long long w) {
  e[++Edge_Num] = {y, head[x], w};
  head[x] = Edge_Num;
}
inline void G_init(int n) {
  memset(head, 0, sizeof(int) * (n + 100));
  Edge_Num = 0;
}
int dir[8][2] = {{-1, 0}, {0, -1}, {-1, -1}, {1, -1},
                 {1, 0},  {0, 1},  {1, 1},   {-1, 1}};
const long double PI = 3.14159265358979323846;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
inline long long rd() {
  long long x = 0;
  bool f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return f ? x : -x;
}
const double eps = 1e-8;
const long long mod = 998244353;
const int M = 1e6 + 10;
const int N = 1e6 + 10;
int n, m, k;
int deg[N];
queue<int> q;
vector<int> ans;
bool vis[N];
bool del[N];
vector<int> ver[N];
bool check(int x, int y) {
  int p = lower_bound(ver[x].begin(), ver[x].end(), y) - ver[x].begin();
  if (p == ver[x].size() || ver[x][p] != y) return 0;
  return 1;
}
vector<int> cliq;
bool check(int x) {
  cliq.clear();
  cliq.push_back(x);
  for (int i = head[x]; i; i = e[i].next) {
    int y = e[i].to;
    if (del[y]) continue;
    cliq.push_back(y);
  }
  assert(cliq.size() == k);
  for (int i = 0; i < cliq.size(); i++) {
    for (int j = i + 1; j < cliq.size(); j++) {
      int x = cliq[i], y = cliq[j];
      if (!check(x, y)) return 0;
    }
  }
  cout << 2 << endl;
  for (auto& x : cliq) cout << x << ' ';
  cout << endl;
  return 1;
}
bool bfs1() {
  while (q.size()) q.pop();
  for (int i = 1; i <= n; i++) {
    sort(ver[i].begin(), ver[i].end());
    if (deg[i] < k) q.push(i);
  }
  while (q.size()) {
    int x = q.front();
    q.pop();
    if (deg[x] == k - 1) {
      if (check(x)) return 1;
    }
    del[x] = 1;
    for (int i = head[x]; i; i = e[i].next) {
      int y = e[i].to;
      if (del[y]) continue;
      if ((--deg[y]) == k - 1) q.push(y);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (deg[i] >= k) {
      ans.push_back(i);
    }
  }
  if (ans.size()) {
    cout << 1 << ' ' << ans.size() << endl;
    for (auto& x : ans) cout << x << ' ';
    cout << endl;
    ans.clear();
    return 1;
  }
  ans.clear();
  return 0;
}
void solve() {
  n = rd(), m = rd(), k = rd();
  G_init(n);
  memset(deg, 0, sizeof(int) * (n + 10));
  memset(del, 0, sizeof(bool) * (n + 10));
  for (int i = 1; i <= n; i++) ver[i].clear();
  for (int i = 1; i <= m; i++) {
    int u = rd(), v = rd();
    ade(u, v, 1);
    ade(v, u, 1);
    ver[u].push_back(v);
    ver[v].push_back(u);
    ++deg[u];
    ++deg[v];
  }
  if (1ll * k * k - k > 2 * m) {
    puts("-1");
    return;
  }
  if (bfs1()) return;
  puts("-1");
}
int main() {
  int T = 1;
  T = rd();
  while (T--) solve();
}