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 head[4010], to[80010], nxt[80010], val[80010], tot = 1, n1, n2, m, S, T,
                                                   dis[4010], deg[4010], t1, t2;
vector<int> ansl[4010];
inline void addedge(int b, int e, int v) {
  nxt[++tot] = head[b];
  head[b] = tot;
  to[tot] = e;
  val[tot] = v;
  nxt[++tot] = head[e];
  head[e] = tot;
  to[tot] = b;
  val[tot] = 0;
}
queue<int> q;
bool bfs() {
  for (int i = S + 1; i <= T; ++i) dis[i] = 0;
  dis[S] = 1;
  q.push(S);
  while (!q.empty()) {
    int t = q.front();
    q.pop();
    for (int i = head[t]; i; i = nxt[i])
      if (val[i] && !dis[to[i]]) {
        q.push(to[i]);
        dis[to[i]] = dis[t] + 1;
      }
  }
  return dis[T] > 0;
}
int dinic(int u, int minv) {
  if (u == T || !minv) return minv;
  int ans = 0, t;
  for (int i = head[u]; i; i = nxt[i])
    if (val[i] && dis[to[i]] == dis[u] + 1 &&
        (t = dinic(to[i], min(minv, val[i])))) {
      val[i] -= t;
      val[i ^ 1] += t;
      minv -= t;
      ans += t;
      if (!minv) break;
    }
  if (!ans) dis[u] = -1;
  return ans;
}
int main() {
  scanf("%d%d%d", &n1, &n2, &m);
  S = 0;
  T = n1 + n2 + 1;
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", &t1, &t2);
    addedge(t1, n1 + t2, 1);
    ++deg[t1];
    ++deg[n1 + t2];
  }
  int mind = 0x3f3f3f3f;
  for (int i = 1; i <= n1 + n2; ++i) mind = min(mind, deg[i]);
  for (int i = 1; i <= n1; ++i) addedge(S, i, deg[i] - mind);
  for (int i = 1; i <= n2; ++i) addedge(n1 + i, T, deg[i + n1] - mind);
  for (int i = mind; i >= 1; --i) {
    while (bfs()) dinic(S, 0x3f3f3f3f);
    for (int j = 2; j < (1 + m << 1); j += 2)
      if (val[j]) ansl[i].push_back(j >> 1);
    for (int j = 1 + m << 1; j <= tot; j += 2) ++val[j];
  }
  for (int i = 0; i <= mind; ++i) {
    printf("%d ", ansl[i].size());
    for (int j = 0; j < ansl[i].size(); ++j) printf("%d ", ansl[i][j]);
    putchar(10);
  }
  return 0;
}