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 inf = 1e9;
const int maxN = 4221;
const int maxM = 15000;
int n, n1, n2, m;
int in[maxN];
vector<int> ans[maxM];
class Graph {
 private:
  int cnt;
  int Head[maxN];
  int Next[maxM];
  int W[maxM];
  int V[maxM];
  int Depth[maxN];
  int cur[maxN];

 public:
  int s, t;
  void init() {
    cnt = -1;
    memset(Head, -1, sizeof(Head));
    memset(Next, -1, sizeof(Next));
  }
  void _Add(int u, int v, int w) {
    cnt++;
    Next[cnt] = Head[u];
    Head[u] = cnt;
    V[cnt] = v;
    W[cnt] = w;
  }
  void Add_Edge(int u, int v, int w) {
    _Add(u, v, w);
    _Add(v, u, 0);
  }
  int dfs(int u, int flow) {
    if (u == t) return flow;
    for (int& i = cur[u]; i != -1; i = Next[i]) {
      if ((Depth[V[i]] == Depth[u] + 1) && (W[i] != 0)) {
        int di = dfs(V[i], min(flow, W[i]));
        if (di > 0) {
          W[i] -= di;
          W[i ^ 1] += di;
          return di;
        }
      }
    }
    return 0;
  }
  int bfs() {
    queue<int> Q;
    while (!Q.empty()) Q.pop();
    memset(Depth, 0, sizeof(Depth));
    Depth[s] = 1;
    Q.push(s);
    do {
      int u = Q.front();
      Q.pop();
      for (int i = Head[u]; i != -1; i = Next[i]) {
        if ((Depth[V[i]] == 0) && (W[i] > 0)) {
          Depth[V[i]] = Depth[u] + 1;
          Q.push(V[i]);
        }
      }
    } while (!Q.empty());
    if (Depth[t] > 0) return 1;
    return 0;
  }
  int Dinic() {
    int Ans = 0;
    while (bfs()) {
      for (int i = 0; i <= n; i++) cur[i] = Head[i];
      while (int d = dfs(s, inf)) {
        Ans += d;
      }
    }
    return Ans;
  }
  void update(int now) {
    for (int i = 1; i <= 2 * m - 1; i += 2) {
      if (W[i] == 0) ans[now].push_back(i / 2 + 1);
    }
    for (int i = 2 * m; i <= 2 * m + 2 * n1 - 1; i += 2) {
      W[i]++;
    }
    for (int i = 2 * m + 2 * n1; i <= 2 * m + 2 * n1 + 2 * n2 - 1; i += 2)
      W[i]++;
  }
} P;
int main() {
  P.init();
  scanf("%d%d%d", &n1, &n2, &m);
  n = n1 + n2 + 1;
  int u, v, mins = inf;
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &u, &v);
    v += n1;
    P.Add_Edge(u, v, 1);
    in[u]++;
    in[v]++;
  }
  for (int i = 1; i < n; i++) mins = min(mins, in[i]);
  for (int i = 1; i <= n1; i++) P.Add_Edge(0, i, in[i] - mins);
  for (int i = 1; i <= n2; i++) P.Add_Edge(i + n1, n, in[i + n1] - mins);
  P.s = 0;
  P.t = n;
  for (int k = 0; k <= mins; k++) {
    P.Dinic();
    P.update(k);
  }
  for (int i = mins; i >= 0; i--) {
    printf("%d", ans[i].size());
    for (int j = 0; j < ans[i].size(); j++) printf(" %d", ans[i][j]);
    printf("\n");
  }
  return 0;
}