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>
#pragma GCC optimize(2)
using namespace std;
const long long inff = 0x3f3f3f3f3f3f3f3f;
const int N = 1e5 + 8;
int n, m, t, k, x, y, du[N], vis[N], sz[N], ans[N];
list<int> g[N];
pair<int, int> a[N];
unordered_map<int, int> mp[N];
int main() {
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    scanf("%d", &n), scanf("%d", &m), scanf("%d", &k);
    for (int i(1); i <= (n); ++i) g[i].clear(), vis[i] = du[i] = 0;
    queue<int> q;
    for (int i(1); i <= (m); ++i) {
      scanf("%d", &x), scanf("%d", &y);
      du[x]++, du[y]++;
      g[x].push_back(y);
      g[y].push_back(x);
      a[i].first = x, a[i].second = y;
    }
    int num = 0;
    for (int i(1); i <= (n); ++i)
      if (du[i] < k) q.push(i), vis[i] = 1, num++;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      for (auto it : g[u]) {
        int v = it;
        if (vis[v]) continue;
        du[v]--;
        if (du[v] < k) q.push(v), vis[v] = 1, num++;
      }
    }
    if (num != n) {
      int ans = 0;
      for (int i(1); i <= (n); ++i) ans += (vis[i] == 0);
      printf("1 %d\n", ans);
      for (int i(1); i <= (n); ++i)
        if (!vis[i]) printf("%d ", i);
      puts("");
      continue;
    }
    if (1ll * k * (k - 1) / 2 > m) {
      puts("-1");
      continue;
    }
    --k;
    for (int i(1); i <= (n); ++i) vis[i] = du[i] = 0;
    for (int i(1); i <= (m); ++i) du[a[i].first]++, du[a[i].second]++;
    for (int i(1); i <= (n); ++i)
      if (du[i] < k) q.push(i), vis[i] = 1, num++;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      for (auto it : g[u]) {
        int v = it;
        if (vis[v]) continue;
        du[v]--;
        if (du[v] < k) q.push(v), vis[v] = 1, num++;
      }
    }
    for (int i(1); i <= (n); ++i) g[i].clear(), du[i] = 0, mp[i].clear();
    for (int i(1); i <= (m); ++i) {
      x = a[i].first, y = a[i].second;
      if (vis[x] || vis[y]) continue;
      g[x].push_back(y);
      g[y].push_back(x);
      mp[x][y] = mp[y][x] = 1;
      du[x]++, du[y]++;
    }
    for (int i(1); i <= (n); ++i) vis[i] = 0;
    for (int i(1); i <= (n); ++i)
      if (du[i] == k) q.push(i), vis[i] = 1;
    int flag = 0;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      if (du[u] == k) flag = 1, num = 0, ans[++num] = u;
      for (auto it : g[u]) {
        int v = it;
        du[v]--;
        if (flag && vis[v] != 2) ans[++num] = v;
        if (vis[v]) continue;
        if (du[v] == k) q.push(v), vis[v] = 1;
      }
      srand(time(0));
      vis[u] = 2;
      if (flag) {
        for (int i(1); i <= (num); ++i)
          if (du[ans[i]] < k - 1) flag = 0;
        if (!flag) continue;
        random_shuffle(ans + 1, ans + num + 1);
        for (int i(1); i <= (num); ++i) {
          for (int j(i + 1); j <= (num); ++j)
            if (mp[ans[i]].find(ans[j]) == mp[ans[i]].end()) {
              flag = 0;
              break;
            }
          if (!flag) break;
        }
        if (flag) break;
      }
    }
    while (!q.empty()) q.pop();
    if (flag) {
      puts("2");
      for (int i(1); i <= (num - 1); ++i) printf("%d ", ans[i]);
      printf("%d\n", ans[num]);
    } else
      puts("-1");
  }
  return 0;
}