Problem: You are given a bipartite graph G = (U, V, E), U is the set of vertices of the first part, V is the set of vertices of the second part and E is the set of edges. There might be multiple edges.

Let's call some subset of its edges <image> k-covering iff the graph <image> has each of its vertices incident to at least k edges. Minimal k-covering is such a k-covering that the size of the subset <image> is minimal possible.

Your task is to find minimal k-covering for each <image>, where minDegree is the minimal degree of any vertex in graph G.

Input

The first line contains three integers n1, n2 and m (1 ≤ n1, n2 ≤ 2000, 0 ≤ m ≤ 2000) — the number of vertices in the first part, the number of vertices in the second part and the number of edges, respectively.

The i-th of the next m lines contain two integers ui and vi (1 ≤ ui ≤ n1, 1 ≤ vi ≤ n2) — the description of the i-th edge, ui is the index of the vertex in the first part and vi is the index of the vertex in the second part.

Output

For each <image> print the subset of edges (minimal k-covering) in separate line.

The first integer cntk of the k-th line is the number of edges in minimal k-covering of the graph. Then cntk integers follow — original indices of the edges which belong to the minimal k-covering, these indices should be pairwise distinct. Edges are numbered 1 through m in order they are given in the input.

Examples

Input

3 3 7
1 2
2 3
1 3
3 2
3 3
2 1
2 1


Output

0 
3 3 7 4 
6 1 3 6 7 4 5 


Input

1 1 5
1 1
1 1
1 1
1 1
1 1


Output

0 
1 5 
2 4 5 
3 3 4 5 
4 2 3 4 5 
5 1 2 3 4 5 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 4007;
const int MX = 1e9 + 7;
const long long int INF = 1e18 + 9LL;
int n, n1, n2, m;
vector<pair<int, int> > G[N];
int d[N];
pair<int, int> p[N];
bool res[N];
int flow[N + N];
queue<int> Q;
bool bfs(int s, int t) {
  for (int i = 1; i <= n; ++i) d[i] = -1;
  d[s] = 0;
  Q.push(s);
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (pair<int, int> v : G[u])
      if (d[v.first] == -1 && flow[v.second]) {
        d[v.first] = d[u] + 1;
        p[v.first] = {u, v.second};
        Q.push(v.first);
      }
  }
  return d[t] != -1;
}
void maxFlow() {
  while (bfs(0, n)) {
    int cur = n;
    while (cur != 0) {
      pair<int, int> t = p[cur];
      flow[t.second] -= 1;
      if (t.second <= m)
        flow[t.second + m] += 1;
      else if (t.second <= m + m)
        flow[t.second - m] += 1;
      cur = t.first;
    }
  }
}
int main() {
  scanf("%d %d %d", &n1, &n2, &m);
  n = n1 + n2 + 1;
  for (int i = 1; i <= m; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    flow[i] = 1;
    G[u].push_back({v + n1, i});
    G[v + n1].push_back({u, i + m});
  }
  int turns = m;
  for (int i = 1; i <= n1; ++i) {
    G[0].push_back({i, 2 * m + i});
    turns = min(turns, (int)G[i].size());
  }
  for (int i = n1 + 1; i < n; ++i) {
    G[i].push_back({n, 2 * m + i});
    turns = min(turns, (int)G[i].size() - 1);
  }
  puts("0");
  for (int i = 1; i <= turns; ++i) {
    for (int j = 1; j <= m; ++j) res[j] = false;
    for (int j = 2 * m + 1; j < 2 * m + n; ++j) ++flow[j];
    maxFlow();
    for (int j = 1; j <= m; ++j)
      if (flow[j] == 0) res[j] = true;
    for (int j = 1; j < n; ++j) {
      int need = flow[2 * m + j];
      for (pair<int, int> v : G[j]) {
        int type = v.second > m ? v.second - m : v.second;
        if (!res[type] && need) {
          --need;
          res[type] = true;
        }
      }
    }
    int ile = 0;
    for (int j = 1; j <= m; ++j)
      if (res[j]) ++ile;
    printf("%d ", ile);
    for (int j = 1; j <= m; ++j)
      if (res[j]) printf("%d ", j);
    puts("");
  }
  return 0;
}