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;
struct MF {
  static const int mxn = 4005;
  struct E {
    int v, w, i, r;
    E(int v, int w, int i, int r) : v(v), w(w), i(i), r(r) {}
  };
  vector<E> g[mxn];
  inline void add_edge(int u, int v, int w, int i) {
    g[u].push_back(E(v, w, i, (int)(g[v].size())));
    g[v].push_back(E(u, 0, i, (int)(g[u].size()) - 1));
  }
  int dis[mxn], que[mxn], ql, qr;
  bool bfs(int S, int T) {
    memset(dis, -1, sizeof(dis));
    ql = qr = 0;
    dis[S] = 0;
    que[qr++] = S;
    while (ql < qr) {
      int u = que[ql++];
      if (u == T) break;
      for (int i = 0; i < (int)(g[u].size()); ++i) {
        int v = g[u][i].v, w = g[u][i].w;
        if (w && dis[v] == -1) {
          dis[v] = dis[u] + 1;
          que[qr++] = v;
        }
      }
    }
    return dis[T] != -1;
  }
  int id[mxn];
  int dfs(int u, int T, int f) {
    if (u == T) return f;
    int flow = 0;
    for (int& i = id[u]; i < (int)(g[u].size()); ++i) {
      int v = g[u][i].v, w = g[u][i].w;
      if (w && dis[v] == dis[u] + 1) {
        int d = dfs(v, T, min(f, w));
        if (d > 0) {
          f -= d;
          flow += d;
          g[u][i].w -= d;
          g[v][g[u][i].r].w += d;
          if (f == 0) break;
        }
      }
    }
    return flow;
  }
  int solve(int S, int T) {
    int ans = 0;
    while (bfs(S, T)) {
      memset(id, 0, sizeof(id));
      ans += dfs(S, T, 0x3f3f3f3f);
    }
    return ans;
  }
} mf;
int n1, n2, m;
int deg[4005];
vector<int> Ans[2005];
int main() {
  scanf("%d %d %d", &n1, &n2, &m);
  int S = n1 + n2, T = S + 1;
  for (int i = 0; i < (int)(m); ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    --u;
    --v;
    ++deg[u];
    ++deg[v + n1];
    mf.add_edge(u, v + n1, 1, i);
  }
  int mn = 0x3f3f3f3f;
  for (int i = 0; i < (int)(n1 + n2); ++i) mn = min(mn, deg[i]);
  for (int i = 0; i < (int)(n1); ++i) mf.add_edge(S, i, deg[i] - mn, -1);
  for (int i = 0; i < (int)(n2); ++i)
    mf.add_edge(i + n1, T, deg[i + n1] - mn, -1);
  for (int k = mn; k >= 0; --k) {
    mf.solve(S, T);
    vector<int> ans;
    for (int i = 0; i < (int)(n1); ++i)
      for (int j = 0; j < (int)(mf.g[i].size()); ++j)
        if (mf.g[i][j].v >= n1 && mf.g[i][j].v < n1 + n2)
          if (mf.g[i][j].w) ans.push_back(mf.g[i][j].i);
    Ans[k] = ans;
    for (int i = 0; i < (int)(n1); ++i) mf.add_edge(S, i, 1, -1);
    for (int i = 0; i < (int)(n2); ++i) mf.add_edge(i + n1, T, 1, -1);
  }
  for (int k = 0; k < (int)(mn + 1); ++k) {
    printf("%d", (int)(Ans[k].size()));
    for (int i = 0; i < (int)(Ans[k].size()); ++i) printf(" %d", Ans[k][i] + 1);
    printf("\n");
  }
  return 0;
}