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 maxn = 4000, inf = 1e9, maxm = 6000;
struct Edge {
  int from, to, cap, flow;
} edge[maxm * 2 + 5];
;
vector<int> g[maxn + 5];
int step[maxn + 5];
int iter[maxn + 5];
int n1, n2, m, S, T, len, mind, start;
int d[maxn + 5];
int res[maxn + 5];
void addedge(int from, int to, int cap) {
  ++len;
  edge[len] = {from, to, cap, 0};
  g[from].push_back(len);
  ++len;
  edge[len] = {to, from, 0, 0};
  g[to].push_back(len);
}
void bfs(int S) {
  memset(step, -1, sizeof(step));
  step[S] = 0;
  queue<int> q;
  q.push(S);
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (int i = 0; i < g[v].size(); ++i) {
      Edge &e = edge[g[v][i]];
      if (e.cap > e.flow && step[e.to] < 0) {
        step[e.to] = step[v] + 1;
        q.push(e.to);
      }
    }
  }
}
int dfs(int v, int t, int f) {
  if (v == t) return f;
  for (int &i = iter[v]; i < g[v].size(); ++i) {
    Edge &e = edge[g[v][i]];
    if (e.cap > e.flow && step[e.to] > step[v]) {
      int d = dfs(e.to, t, min(e.cap - e.flow, f));
      if (d > 0) {
        e.flow += d;
        edge[g[v][i] ^ 1].flow -= d;
        return d;
      }
    }
  }
  return 0;
}
int maxflow(int S, int T) {
  int flow = 0;
  for (;;) {
    bfs(S);
    if (step[T] < 0) return flow;
    memset(iter, 0, sizeof(iter));
    int f;
    while ((f = dfs(S, T, inf)) > 0) flow += f;
  }
}
vector<int> ans[2000 + 5];
int main() {
  scanf("%d%d%d", &n1, &n2, &m);
  S = 0;
  T = n1 + n2 + 1;
  len = -1;
  for (int i = 1; i <= m; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    y += n1;
    ++d[x], ++d[y];
    addedge(x, y, 1);
  }
  mind = d[1];
  for (int i = 2; i <= n1 + n2; ++i) mind = min(mind, d[i]);
  start = len + 1;
  for (int i = 1; i <= n1; ++i) addedge(S, i, 1);
  for (int i = 1; i <= n2; ++i) addedge(i + n1, T, 1);
  printf("0\n");
  for (int i = mind; i >= 1; --i) {
    for (int j = start; j <= len; j += 2)
      edge[j].cap = d[(j - start) / 2 + 1] - i;
    maxflow(S, T);
    for (int j = 0; j < start; j += 2)
      if (edge[j].flow < edge[j].cap) ans[i].push_back(j / 2 + 1);
  }
  for (int i = 1; i <= mind; ++i) {
    printf("%d ", ans[i].size());
    for (auto x : ans[i]) printf("%d ", x);
    printf("\n");
  }
  return 0;
}