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 = 1e5 + 100;
struct node {
  int x, y;
} e[maxn * 2];
struct node2 {
  int u, v, next;
} edge[maxn * 2];
map<pair<int, int>, int> ma;
vector<int> vv[maxn];
vector<int> v2;
queue<int> q;
int head[maxn], cnt, vis[maxn], deg[maxn], k;
void add(int x, int y) {
  edge[cnt].next = head[x];
  edge[cnt].v = y;
  head[x] = cnt++;
}
bool solve(int x) {
  int v;
  node e;
  v2.clear();
  v2.push_back(x);
  for (int i = head[x]; i != -1; i = edge[i].next) {
    v = edge[i].v;
    if (vis[v]) continue;
    v2.push_back(v);
  }
  if (v2.size() != k) return false;
  for (int i = 0; i < v2.size(); i++) {
    for (int j = i + 1; j < v2.size(); j++) {
      int x = v2[i];
      int y = v2[j];
      if (x > y) swap(x, y);
      e.x = x;
      e.y = y;
      if (!binary_search(vv[x].begin(), vv[x].end(), y)) return false;
    }
  }
  return true;
}
int main() {
  int T, v, n, m, u;
  scanf("%d", &T);
  while (T--) {
    cnt = 0;
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= n; i++) {
      head[i] = -1;
      vv[i].clear();
      deg[i] = 0;
      vis[i] = 0;
    }
    for (int i = 1; i <= m; i++) {
      scanf("%d%d", &u, &v);
      if (u > v) swap(u, v);
      e[i].x = u;
      e[i].y = v;
      vv[u].push_back(v);
      add(u, v);
      add(v, u);
      deg[u]++;
      deg[v]++;
    }
    for (int i = 1; i <= n; i++) {
      sort(vv[i].begin(), vv[i].end());
    }
    int left = n;
    for (int i = 1; i <= n; i++) {
      if (deg[i] < k) {
        q.push(i);
        vis[i] = 1;
        left--;
      }
    }
    while (!q.empty()) {
      u = q.front();
      q.pop();
      for (int i = head[u]; i != -1; i = edge[i].next) {
        v = edge[i].v;
        deg[v]--;
        if (deg[v] < k && !vis[v]) {
          q.push(v);
          vis[v] = 1;
          left--;
        }
      }
    }
    if (left) {
      printf("%d %d\n", 1, left);
      for (int i = 1; i <= n; i++) {
        if (!vis[i]) {
          printf("%d ", i);
        }
      }
      printf("\n");
      continue;
    }
    int step = (int)sqrt(2 * m + 2) + 5;
    if (k >= step) {
      printf("-1\n");
      continue;
    }
    for (int i = 1; i <= n; i++) {
      vis[i] = 0;
      deg[i] = 0;
    }
    for (int i = 1; i <= m; i++) {
      u = e[i].x;
      v = e[i].y;
      deg[u]++;
      deg[v]++;
    }
    left = n;
    for (int i = 1; i <= n; i++) {
      if (deg[i] < k - 1) {
        q.push(i);
        vis[i] = 1;
        left--;
      }
    }
    while (!q.empty()) {
      u = q.front();
      q.pop();
      for (int i = head[u]; i != -1; i = edge[i].next) {
        v = edge[i].v;
        deg[v]--;
        if (deg[v] < k - 1 && !vis[v]) {
          q.push(v);
          vis[v] = 1;
          left--;
        }
      }
    }
    for (int i = 1; i <= n; i++) {
      if (!vis[i] && deg[i] == k - 1) {
        q.push(i);
      }
    }
    int flag = 0;
    while (!q.empty()) {
      int i = q.front();
      q.pop();
      if (solve(i)) {
        printf("%d\n", 2);
        for (int j = head[i]; j != -1; j = edge[j].next) {
          v = edge[j].v;
          if (!vis[v]) {
            printf("%d ", v);
          }
        }
        printf("%d\n", i);
        flag = 1;
        break;
      } else {
        vis[i] = 1;
        for (int j = head[i]; j != -1; j = edge[j].next) {
          v = edge[j].v;
          deg[v]--;
          if (!vis[v]) {
            if (deg[v] == k - 1 && !vis[v]) {
              q.push(v);
            }
          }
        }
      }
    }
    while (!q.empty()) q.pop();
    if (flag == 0) {
      printf("-1\n");
    }
  }
  return 0;
}