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 = 5010, inf = 1e9;
int n1, n2, m, md, cnt = 1, S, T, G[N], du[N], cur[N];
struct edge {
  int t, nx, f, g;
} E[N * 20];
inline void addedge(int x, int y, int f, int g = 0) {
  E[++cnt].t = y;
  E[cnt].nx = G[x];
  G[x] = cnt;
  E[cnt].f = f;
  E[cnt].g = g;
  E[++cnt].t = x;
  E[cnt].nx = G[y];
  G[y] = cnt;
  E[cnt].f = 0;
  E[cnt].g = g;
}
vector<int> ans[N];
int dis[N];
queue<int> Q;
inline bool bfs() {
  for (int i = S; i <= T; i++) dis[i] = -1;
  dis[S] = 0;
  Q.push(S);
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    for (int i = G[x]; i; i = E[i].nx)
      if (E[i].f && !~dis[E[i].t]) {
        dis[E[i].t] = dis[x] + 1;
        Q.push(E[i].t);
      }
  }
  return ~dis[T];
}
int dfs(int x, int f) {
  if (x == T || !f) return f;
  int used = 0, w;
  for (int &i = cur[x]; i; i = E[i].nx)
    if (E[i].f && dis[E[i].t] == dis[x] + 1) {
      w = dfs(E[i].t, min(f - used, E[i].f));
      E[i].f -= w;
      E[i ^ 1].f += w;
      if ((used += w) == f) return used;
    }
  if (!used) dis[x] = -1;
  return used;
}
int main() {
  scanf("%d%d%d", &n1, &n2, &m);
  md = 1e9;
  for (int i = 1, x, y; i <= m; i++)
    scanf("%d%d", &x, &y), addedge(x, y + n1, 1, i), du[x]++, du[y + n1]++;
  for (int i = 1; i <= n1 + n2; i++) md = min(md, du[i]);
  T = n1 + n2 + 1;
  for (int i = 1; i <= n1; i++) addedge(S, i, du[i] - md - 1);
  for (int i = 1; i <= n2; i++) addedge(n1 + i, T, du[i + n1] - md - 1);
  for (int k = md; ~k; k--) {
    for (int i = G[S]; i; i = E[i].nx) E[i].f++;
    for (int i = G[T]; i; i = E[i].nx) E[i ^ 1].f++;
    while (bfs()) {
      for (int i = S; i <= T; i++) cur[i] = G[i];
      dfs(S, inf);
    }
    for (int x = 1; x <= n1; x++)
      for (int i = G[x]; i; i = E[i].nx)
        if (E[i].g && E[i].f) ans[k].push_back(E[i].g);
  }
  for (int i = 0; i <= md; i++) {
    printf("%d ", (int)ans[i].size());
    for (int j = 0; j < ans[i].size(); j++) printf("%d ", ans[i][j]);
    putchar('\n');
  }
  return 0;
}