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;
int res[4002][4002], mf, f, s, t, r[4002][4002], d[4000];
int n1, n2;
vector<int> p;
void augment(int v, int minEdge) {
  if (v == s) {
    f = minEdge;
  } else if (p[v] != -1) {
    augment(p[v], min(minEdge, res[p[v]][v]));
    res[p[v]][v] -= f;
    res[v][p[v]] += f;
  }
}
int max_flow() {
  mf = 0;
  int MAX_V = n1 + n2 + 2;
  while (1) {
    f = 0;
    vector<int> dist(MAX_V, 99999999);
    dist[s] = 0;
    queue<int> q;
    q.push(s);
    p.assign(MAX_V, -1);
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      if (u == t) break;
      for (int v = 0; v < MAX_V; v++)
        if (res[u][v] > 0 && dist[v] == 99999999) {
          dist[v] = dist[u] + 1;
          q.push(v);
          p[v] = u;
        }
    }
    augment(t, 99999999);
    if (f == 0) break;
    mf += f;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int m, a[2001], b[2001], md, ea, eb;
  cin >> n1 >> n2 >> m;
  s = n1 + n2;
  t = s + 1;
  for (int i = (0); i < (m); ++i) {
    cin >> ea >> eb;
    a[i] = ea - 1;
    b[i] = eb - 1 + n1;
    d[a[i]]++;
    d[b[i]]++;
    r[a[i]][b[i]]++;
  }
  md = 99999999;
  for (int i = (0); i < (n1 + n2); ++i)
    if (d[i] < md) md = d[i];
  cout << "0\n";
  for (int k = (1); k < (md + 1); ++k) {
    for (int i = (0); i < (n1); ++i) {
      res[s][i] = d[i] - k;
      res[i][t];
    }
    for (int i = (n1); i < (n1 + n2); ++i) {
      res[s][i] = 0;
      res[i][t] = d[i] - k;
    }
    for (int i = (0); i < (n1 + n2); ++i)
      for (int j = (0); j < (n1 + n2); ++j) res[i][j] = r[i][j];
    max_flow();
    cout << m - mf;
    int t;
    for (int i = (0); i < (m); ++i) {
      if (res[a[i]][b[i]] > 0) {
        cout << " " << i + 1;
        res[a[i]][b[i]]--;
      }
    }
    cout << "\n";
  }
}