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 = 1e4 + 5;
int deg[maxn], minc[maxn];
bool used[maxn];
vector<int> ans[maxn];
struct dinic {
  static const int inf = 1e9;
  struct edge {
    int dest, cap, flow, rev, idx;
    bool inc;
    edge(int a, int b, int c, int e, bool d)
        : dest(a), cap(b), rev(c), flow(0), inc(d), idx(e) {}
    int operator()(int offset) const { return inc ? cap + offset : cap; }
  };
  vector<edge> graf[maxn];
  int level[maxn], qu[maxn], l, r;
  void add_edge(int from, int dest, int cap, int idx, bool inc) {
    graf[from].emplace_back(dest, cap, graf[dest].size(), idx, inc);
    graf[dest].emplace_back(from, 0, graf[from].size() - 1, idx, inc);
  }
  bool bfs(int s, int t, int offset) {
    memset(level, -1, sizeof(level));
    level[s] = 0;
    l = r = 0;
    qu[r++] = s;
    while (l < r) {
      int x = qu[l++];
      for (edge &e : graf[x])
        if (e(offset) > e.flow && level[e.dest] == -1) {
          level[e.dest] = level[x] + 1;
          qu[r++] = e.dest;
        }
    }
    return level[t] != -1;
  }
  int dfs(int x, int t, int offset, int low) {
    if (x == t) return low;
    int r = 0;
    for (edge &e : graf[x])
      if (e(offset) > e.flow && level[e.dest] == level[x] + 1) {
        int f = dfs(e.dest, t, offset, min(low - r, e(offset) - e.flow));
        e.flow += f;
        graf[e.dest][e.rev].flow -= f;
        r += f;
      }
    if (r == 0) level[x] = -1;
    return r;
  }
  int operator()(int s, int t, int offset) {
    static int r = 0;
    for (; bfs(s, t, offset); r += dfs(s, t, offset, inf))
      ;
    return r;
  }
} flow;
int main() {
  int n1, n2, m;
  scanf("%d %d %d", &n1, &n2, &m);
  int s = 0, t = n1 + n2 + 1;
  for (int i = 0; i < m; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    flow.add_edge(a, b + n1, 1, i, false);
    ++deg[a], ++deg[b + n1];
  }
  int mindeg = maxn;
  for (int i = 1; i <= n1 + n2; ++i) mindeg = min(mindeg, deg[i]);
  for (int i = 1; i <= n1; ++i) flow.add_edge(s, i, deg[i] - mindeg, -1, true);
  for (int i = 1; i <= n2; ++i)
    flow.add_edge(i + n1, t, deg[i + n1] - mindeg, -1, true);
  for (int i = 0; i <= mindeg; ++i) {
    int f = flow(s, t, i);
    minc[mindeg - i] = m - f;
    memset(used, false, sizeof(used));
    for (int j = 1; j <= n1; ++j) {
      for (auto e : flow.graf[j])
        if (e.dest > n1 && e.dest <= n1 + n2) {
          if (e.flow == 1) used[e.idx] = true;
        }
    }
    for (int j = 0; j < m; ++j)
      if (!used[j]) ans[mindeg - i].push_back(j);
  }
  for (int i = 0; i <= mindeg; ++i) {
    assert(minc[i] == (int)ans[i].size());
    printf("%d ", minc[i]);
    for (int j = 0; j < ans[i].size(); ++j) printf("%d ", ans[i][j] + 1);
    puts("");
  }
  return 0;
}