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>
const int N = 2005;
const int inf = 1000000000;
int n1, n2, m, s, t, num[N * 2], cnt, last[N * 2], cur[N * 2], dis[N * 2],
    tot[N], ans[N][N], deg[N * 2];
struct edge {
  int to, c, next;
} e[N * 10];
std::queue<int> que;
void addedge(int u, int v, int c) {
  e[++cnt].to = v;
  e[cnt].c = c;
  e[cnt].next = last[u];
  last[u] = cnt;
  e[++cnt].to = u;
  e[cnt].c = 0;
  e[cnt].next = last[v];
  last[v] = cnt;
}
bool bfs() {
  for (int i = s; i <= t; i++) dis[i] = 0;
  while (!que.empty()) que.pop();
  dis[s] = 1;
  que.push(s);
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    for (int i = last[u]; i; i = e[i].next)
      if (e[i].c && !dis[e[i].to]) {
        dis[e[i].to] = dis[u] + 1;
        if (e[i].to == t) return 1;
        que.push(e[i].to);
      }
  }
  return 0;
}
int dfs(int x, int maxf) {
  if (x == t || !maxf) return maxf;
  int ret = 0;
  for (int &i = cur[x]; i; i = e[i].next)
    if (e[i].c && dis[e[i].to] == dis[x] + 1) {
      int f = dfs(e[i].to, std::min(e[i].c, maxf - ret));
      e[i].c -= f;
      e[i ^ 1].c += f;
      ret += f;
      if (maxf == ret) break;
    }
  return ret;
}
void dinic() {
  while (bfs()) {
    for (int i = s; i <= t; i++) cur[i] = last[i];
    dfs(s, inf);
  }
}
int main() {
  scanf("%d%d%d", &n1, &n2, &m);
  s = 0;
  t = n1 + n2 + 1;
  cnt = 1;
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    deg[x]++;
    deg[y + n1]++;
    addedge(x, y + n1, 1);
  }
  int mn = m, tmp = cnt;
  for (int i = 1; i <= n1 + n2; i++) mn = std::min(mn, deg[i]);
  for (int i = 1; i <= n1; i++) addedge(s, i, deg[i] - mn), num[i] = cnt;
  for (int i = n1 + 1; i <= n1 + n2; i++)
    addedge(i, t, deg[i] - mn), num[i] = cnt;
  for (int i = mn; i >= 0; i--) {
    dinic();
    for (int j = 2; j <= tmp; j += 2)
      if (e[j].c) ans[i][++tot[i]] = j / 2;
    for (int j = 1; j <= n1 + n2; j++) e[num[j] ^ 1].c++;
  }
  for (int i = 0; i <= mn; i++) {
    printf("%d ", tot[i]);
    for (int j = 1; j <= tot[i]; j++) printf("%d ", ans[i][j]);
    puts("");
  }
  return 0;
}