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;
int n1, n2, m, tmp, res, S, T, d[4005], po[4005], pos[4005];
int k = 1, la[4005], ff[12005], q[4005], dep[4005], cur[4005];
struct node {
  int a, b, c;
} e[12005];
vector<int> ans[4005];
void add(int a, int b, int c) {
  e[++k] = (node){a, b, c};
  ff[k] = la[a];
  la[a] = k;
  e[++k] = (node){b, a, 0};
  ff[k] = la[b];
  la[b] = k;
}
int dfs(int x, int flow) {
  if (x == T) return flow;
  int res = 0, tmp;
  for (int a = cur[x]; a && flow; a = ff[a]) {
    cur[x] = a;
    if (e[a].c && dep[e[a].b] == dep[x] + 1) {
      tmp = dfs(e[a].b, min(flow, e[a].c));
      res += tmp;
      flow -= tmp;
      e[a].c -= tmp;
      e[a ^ 1].c += tmp;
    }
  }
  if (!res) dep[x] = -1;
  return res;
}
bool bfs() {
  for (int i = S; i <= T; i++) dep[i] = 0, cur[i] = la[i];
  int l = 1, r = 2;
  q[1] = S;
  dep[S] = 1;
  while (l < r) {
    int x = q[l++];
    for (int a = la[x]; a; a = ff[a])
      if (e[a].c && !dep[e[a].b]) q[r] = e[a].b, dep[q[r]] = dep[x] + 1, r++;
  }
  return dep[T];
}
int main() {
  int a, b;
  scanf("%d%d%d", &n1, &n2, &m);
  S = 0;
  T = n1 + n2 + 1;
  tmp = m;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a, &b);
    add(a, n1 + b, 1);
    d[a]++;
    d[n1 + b]++;
    po[i] = k - 1;
  }
  for (int i = 1; i <= n1 + n2; i++) tmp = min(tmp, d[i]);
  for (int i = 1; i <= n1; i++) add(S, i, d[i] - tmp), pos[i] = k - 1;
  for (int i = n1 + 1; i <= n1 + n2; i++) add(i, T, d[i] - tmp), pos[i] = k - 1;
  for (int j = tmp; j; j--) {
    while (bfs()) res += dfs(S, 1050000000);
    for (int i = 1; i <= m; i++)
      if (e[po[i]].c) ans[j].push_back(i);
    for (int i = 1; i <= n1 + n2; i++) e[pos[i]].c++;
  }
  for (int i = 0; i <= tmp; i++) {
    printf("%d", ans[i].size());
    for (int j = 0; j < ans[i].size(); j++) printf(" %d", ans[i][j]);
    printf("\n");
  }
  return 0;
}