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 N = (int)2000 + 10;
int n1, n2, m, s, t;
int cnt = 1, lst[N * 2], to[N * 10], nxt[N * 10], f[N * 10];
int deg[N * 2], ans[N][N];
int indx, tag[N * 2];
void add(int u, int v, int flow) {
  nxt[++cnt] = lst[u];
  lst[u] = cnt;
  to[cnt] = v;
  f[cnt] = flow;
  nxt[++cnt] = lst[v];
  lst[v] = cnt;
  to[cnt] = u;
  f[cnt] = 0;
}
int dfs(int u) {
  if (u == t) return 1;
  tag[u] = indx;
  for (int j = lst[u]; j; j = nxt[j]) {
    int v = to[j];
    if (!f[j] || tag[v] == indx) continue;
    if (dfs(v)) {
      f[j]--;
      f[j ^ 1]++;
      return 1;
    }
  }
  return 0;
}
int main() {
  scanf("%d%d%d", &n1, &n2, &m);
  for (int i = 1, u, v; i <= m; i++) {
    scanf("%d %d", &u, &v);
    add(u, n1 + v, 1);
    deg[u]++;
    deg[n1 + v]++;
  }
  int T = deg[1];
  for (int i = 2; i <= n1 + n2; i++) T = min(T, deg[i]);
  s = n1 + n2 + 1, t = s + 1;
  for (int i = 1; i <= n1; i++) add(s, i, deg[i] - T);
  for (int i = 1; i <= n2; i++) {
    add(n1 + i, t, deg[n1 + i] - T);
  }
  for (int cvr = T; cvr >= 0; cvr--) {
    indx++;
    while (dfs(s)) indx++;
    for (int i = 1; i <= m; i++)
      if (f[2 * i]) ans[cvr][++ans[cvr][0]] = i;
    for (int i = 1; i <= n1; i++) f[2 * (m + i)]++;
    for (int i = 1; i <= n2; i++) f[2 * (m + n1 + i)]++;
  }
  for (int i = 0; i <= T; i++) {
    printf("%d ", ans[i][0]);
    for (int j = 1; j <= ans[i][0]; j++) printf("%d ", ans[i][j]);
    puts("");
  }
  return 0;
}