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 = 4444;
struct edge {
  int from, to, rflow, nxt, id;
  edge(int from, int to, int rflow, int nxt, int id)
      : from(from), to(to), rflow(rflow), nxt(nxt), id(id) {}
};
int s, t;
int deg[N], last[N], last0[N];
vector<edge> g;
void eadd(int u, int v, int id = 2018, int cap = 1) {
  ++deg[u], ++deg[v];
  g.emplace_back(u, v, cap, last[u], id);
  last[u] = g.size() - 1;
  g.emplace_back(v, u, 0, last[v], -id);
  last[v] = g.size() - 1;
}
int layer[N];
bool bfs() {
  fill(layer, layer + t + 1, -1);
  queue<int> q;
  q.push(s);
  layer[s] = 0;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int e = last[u]; ~e; e = g[e].nxt) {
      edge& ed = g[e];
      if (layer[ed.to] == -1 && ed.rflow > 0) {
        q.push(ed.to);
        layer[ed.to] = layer[u] + 1;
      }
    }
  }
  return ~layer[t];
}
int dfs(int u, int flow) {
  if (u == t) return flow;
  for (int e = last0[u]; ~e; e = g[e].nxt) {
    edge& ed = g[e];
    edge& edr = g[e ^ 1];
    last0[u] = e;
    if (layer[ed.to] == layer[u] + 1 && ed.rflow > 0) {
      int by = dfs(ed.to, min(flow, ed.rflow));
      if (by > 0) {
        ed.rflow -= by;
        edr.rflow += by;
        return by;
      }
    }
  }
  return 0;
}
int mf() {
  int mflow = 0, amt;
  while (bfs()) {
    for (int i = 0; i <= t; ++i) last0[i] = last[i];
    while ((amt = dfs(s, 1e9)) > 0) mflow += amt;
  }
  return mflow;
}
int sat[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie();
  memset(last, -1, sizeof last);
  int n1, n2, m;
  cin >> n1 >> n2 >> m;
  for (int i = 1; i <= m; ++i) {
    int u, v;
    cin >> u >> v;
    eadd(u, n1 + v, i);
  }
  int mn = m;
  for (int i = 1; i <= n1 + n2; ++i) mn = min(mn, deg[i]);
  s = 0, t = n1 + n2 + 1;
  for (int i = 1; i <= n1; ++i) eadd(s, i, 2018, 0);
  for (int i = n1 + 1; i <= n1 + n2; ++i) eadd(i, t, 2018, 0);
  cout << "0\n";
  for (int k = 1; k <= mn; ++k) {
    for (int e = last[s]; ~e; e = g[e].nxt) ++g[e].rflow;
    for (int e = last[t]; ~e; e = g[e].nxt) ++g[e ^ 1].rflow;
    int mflow = mf();
    for (int i = 1; i <= n1 + n2; ++i) sat[i] = 0;
    vector<int> res;
    for (int i = 1; i <= n1; ++i) {
      for (int e = last[i]; ~e; e = g[e].nxt)
        if (g[e].to != s && g[e].rflow == 0) {
          ++sat[g[e].to];
          ++sat[g[e].from];
          res.push_back(g[e].id);
        }
    }
    for (int i = 1; i <= n1; ++i) {
      if (sat[i] < k)
        for (int e = last[i]; ~e && sat[i] < k; e = g[e].nxt)
          if (g[e].to != s && g[e].rflow > 0)
            res.push_back(g[e].id), ++sat[g[e].to], ++sat[i];
    }
    for (int i = n1 + 1; i <= n1 + n2; ++i) {
      if (sat[i] < k) {
        for (int e = last[i]; ~e && sat[i] < k; e = g[e].nxt)
          if (g[e].to != t && g[e ^ 1].rflow > 0)
            res.push_back(g[e ^ 1].id), ++sat[i], ++sat[g[e].to];
      }
    }
    cout << res.size();
    for (int e : res) cout << ' ' << e;
    cout << '\n';
  }
  return 0;
}