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 double PI = 4 * atan(1);
int N, N1, N2, M;
unordered_map<int, int> cap[4013];
map<pair<int, int>, vector<int> > to_idx;
int deg[4013];
vector<int> ans[4013];
bool vis[4013];
bool dfs(int n) {
  if (n == N - 1) return true;
  if (vis[n]) return false;
  vis[n] = true;
  for (auto p : cap[n])
    if (p.second) {
      int v = p.first;
      if (dfs(v)) {
        --cap[n][v];
        ++cap[v][n];
        return true;
      }
    }
  return false;
}
bool augment() {
  memset(vis, 0, sizeof vis);
  return dfs(0);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N1 >> N2 >> M;
  N = N1 + N2 + 2;
  for (int i = 0; i < M; i++) {
    int a, b;
    cin >> a >> b;
    b += N1;
    to_idx[{a, b}].push_back(i + 1);
    cap[a][b] += 1;
    ++deg[a], ++deg[b];
  }
  int minDeg = M;
  for (int i = 1; i < N - 1; i++) minDeg = min(minDeg, deg[i]);
  for (int i = 1; i <= N1; i++) cap[0][i] = deg[i] - minDeg;
  for (int i = N1 + 1; i <= N1 + N2; i++) cap[i][N - 1] = deg[i] - minDeg;
  for (int k = minDeg; k >= 0; k--) {
    while (augment())
      ;
    for (int i = 1; i <= N1; i++) {
      for (auto p : cap[i]) {
        if (p.first != 0)
          for (int j = 0; j < p.second; j++)
            ans[k].push_back(to_idx[{i, p.first}][j]);
      }
    }
    for (int i = 1; i <= N1; i++) ++cap[0][i];
    for (int i = N1 + 1; i <= N1 + N2; i++) ++cap[i][N - 1];
  }
  for (int k = 0; k <= minDeg; k++) {
    cout << ans[k].size();
    for (int x : ans[k]) cout << ' ' << x;
    cout << '\n';
  }
}