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 maxn = 5000;
const double INF = 1e18;
const double ep = 1e-8;
struct Edge {
  int from, to;
  double cap, flow;
  int id;
  Edge() {}
  Edge(int u, int v, double c, double f, int id_)
      : from(u), to(v), cap(c), flow(f), id(id_) {}
};
struct Dinic {
  int n, m, s, t;
  vector<Edge> edges;
  vector<int> G[maxn];
  int d[maxn], cur[maxn];
  bool vis[maxn];
  void init(int n) {
    for (int i = 0; i < maxn; i++) G[i].clear();
    edges.clear();
  }
  void add(int from, int to, double cap, int id) {
    edges.push_back(Edge(from, to, cap, 0, id));
    edges.push_back(Edge(to, from, 0, 0, id));
    m = edges.size();
    G[from].push_back(m - 2);
    G[to].push_back(m - 1);
  }
  bool BFS() {
    memset(vis, 0, sizeof(vis));
    queue<int> Q;
    Q.push(s);
    d[s] = 0;
    vis[s] = 1;
    while (!Q.empty()) {
      int x = Q.front();
      Q.pop();
      for (int i = 0; i < G[x].size(); i++) {
        Edge& e = edges[G[x][i]];
        if (!vis[e.to] && e.cap - e.flow > ep) {
          vis[e.to] = 1;
          d[e.to] = d[x] + 1;
          Q.push(e.to);
        }
      }
    }
    return vis[t];
  }
  double DFS(int x, double a) {
    if (x == t || a == 0) return a;
    double flow = 0, f;
    for (int& i = cur[x]; i < G[x].size(); i++) {
      Edge& e = edges[G[x][i]];
      if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) {
        e.flow += f;
        edges[G[x][i] ^ 1].flow -= f;
        flow += f;
        a -= f;
        if (a == 0) break;
      }
    }
    return flow;
  }
  double maxFlow(int s, int t) {
    this->s = s;
    this->t = t;
    double flow = 0;
    while (BFS()) {
      memset(cur, 0, sizeof(cur));
      flow += DFS(s, INF);
    }
    return flow;
  }
  void show() {
    for (auto& e : edges) {
      printf("(%d -> %d) => cap = %f , flow = %f\n", e.from, e.to, e.cap,
             e.flow);
    }
  }
  void DFS2(int v) {
    vis[v] = true;
    for (int i = 0; i < G[v].size(); ++i) {
      Edge e = edges[G[v][i]];
      if (!vis[e.to] && e.cap - e.flow > ep) {
        DFS2(e.to);
      }
    }
  }
  void showMinCut(int v) {
    DFS2(v);
    for (auto& e : edges) {
      if (vis[e.from] && !vis[e.to] && e.cap != 0 && e.flow != 0)
        printf("(%d -> %d) => cap = %f , flow = %f\n", e.from, e.to, e.cap,
               e.flow);
    }
  }
};
int N1, N2, M;
vector<pair<int, int>> g[2005];
bool vis[2005];
int deg[4005];
map<int, map<int, int>> mp;
int s, t;
void solve(int K) {
  for (int i = 1; i <= M; ++i) vis[i] = 0;
  Dinic G;
  G.init(N1 + N2 + 10);
  for (int i = 1; i <= N1; ++i) {
    G.add(s, i, deg[i] - K, -1);
    for (auto& e : g[i]) {
      G.add(i, e.first, 1, e.second);
    }
  }
  for (int i = N1 + 1; i <= N2 + N1; ++i) {
    G.add(i, t, deg[i] - K, -1);
  }
  long long ret = G.maxFlow(s, t);
  printf("%lld ", M - ret);
  for (auto& e : G.edges) {
    if (e.flow > 0 && e.cap - e.flow <= ep) {
      if (e.from >= 1 && e.from <= N1 && e.to >= N1 + 1 && e.to <= N1 + N2) {
        vis[e.id] = 1;
      }
    }
  }
  for (int i = 1; i <= M; ++i) {
    if (!vis[i]) printf("%d ", i);
  }
  printf("\n");
}
int main() {
  scanf("%d %d %d", &N1, &N2, &M);
  for (int i = 1; i <= M; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    g[x].emplace_back(y + N1, i);
    ++deg[x];
    ++deg[y + N1];
  }
  int min_deg = M + 10;
  for (int i = 1; i <= N1 + N2; ++i) {
    min_deg = min(min_deg, deg[i]);
  }
  s = N1 + N2 + 10 - 2, t = s + 1;
  printf("0\n");
  for (int k = 1; k <= min_deg; ++k) {
    solve(k);
  }
}