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 to[20005], head[20005], nex[20005], u = 1, f[20005], poz[20005];
int s, t;
int f1[20005], f2[20005];
vector<int> ans[20005];
int lvl[20005], last[20005];
queue<int> q;
void add(int x, int y, int nr, int p = 0) {
  to[++u] = y;
  nex[u] = head[x];
  head[x] = u;
  f[u] = nr;
  poz[u] = p;
}
int bfs() {
  int i;
  for (i = 1; i <= t; i++) lvl[i] = -1;
  lvl[s] = 0;
  q.push(s);
  while (!q.empty()) {
    int nod = q.front();
    q.pop();
    for (i = head[nod]; i; i = nex[i])
      if (lvl[to[i]] == -1 && f[i] != 0)
        lvl[to[i]] = lvl[nod] + 1, q.push(to[i]);
  }
  return lvl[t] != -1;
}
int dfs(int nod, int flo) {
  if (flo == 0) return 0;
  if (nod == t) return flo;
  int i;
  for (i = last[nod]; i; i = nex[i]) {
    if (f[i] == 0 || lvl[to[i]] != lvl[nod] + 1) continue;
    int aux = dfs(to[i], min(flo, f[i]));
    if (aux == 0) continue;
    f[i] -= aux;
    f[i ^ 1] += aux;
    last[nod] = i;
    return aux;
  }
  last[nod] = i;
  return 0;
}
int main() {
  int n1, n2, m, i, j, k;
  scanf("%d%d%d", &n1, &n2, &m);
  s = n1 + n2 + 1;
  t = n1 + n2 + 2;
  for (i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, n1 + y, 1, i);
    add(n1 + y, x, 0);
    f1[x]++;
    f2[y]++;
  }
  int mini = 2e9;
  for (i = 1; i <= n1; i++) mini = min(mini, f1[i]);
  for (i = 1; i <= n2; i++) mini = min(mini, f2[i]);
  for (i = 1; i <= n1; i++) add(s, i, f1[i] - mini), add(i, s, 0);
  for (i = 1; i <= n2; i++) add(n1 + i, t, f2[i] - mini), add(t, n1 + i, 0);
  for (k = mini; k >= 0; k--) {
    while (bfs()) {
      for (i = 1; i <= t; i++) last[i] = head[i];
      dfs(s, 2e9);
    }
    for (i = 1; i <= n1; i++)
      for (j = head[i]; j; j = nex[j])
        if (f[j] != 0 && to[j] != s) ans[k].push_back(poz[j]);
    for (i = head[s]; i; i = nex[i]) f[i]++;
    for (i = head[t]; i; i = nex[i]) f[i ^ 1]++;
  }
  for (k = 0; k <= mini; k++) {
    printf("%d ", ans[k].size());
    for (auto it : ans[k]) printf("%d ", it);
    printf("\n");
  }
  return 0;
}