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 = 2005;
const int inf = N;
int n1, n2, m, tot, S, T, ed = 1, fir[N * 2], deg[N * 2], h[N * 2], gap[N * 2],
                          pos[N * 2];
vector<int> ans[N];
struct E {
  int v, w, nxt;
} e[N * 6];
inline void adde(int x, int y, int z) {
  e[++ed].v = y;
  e[ed].w = z;
  e[ed].nxt = fir[x];
  fir[x] = ed;
  e[++ed].v = x;
  e[ed].w = 0;
  e[ed].nxt = fir[y];
  fir[y] = ed;
}
int sap(int x, int flow) {
  if (x == T) return flow;
  int i, minh = tot, dec = 0, tmp;
  for (i = fir[x]; i; i = e[i].nxt)
    if (e[i].w) {
      if (h[x] == h[e[i].v] + 1) {
        tmp = sap(e[i].v, min(flow - dec, e[i].w));
        e[i].w -= tmp;
        e[i ^ 1].w += tmp;
        dec += tmp;
        if (h[S] >= tot) return dec;
        if (dec == flow) return flow;
      }
      if (h[e[i].v] < minh) minh = h[e[i].v];
    }
  if (!dec) {
    if (!(--gap[h[x]])) h[S] = tot;
    ++gap[h[x] = minh + 1];
  }
  return dec;
}
int main() {
  int i, j, k;
  scanf("%d%d%d", &n1, &n2, &m);
  tot = n1 + n2;
  S = ++tot;
  T = ++tot;
  for (i = 1; i <= m; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    y += n1;
    adde(x, y, 1);
    ++deg[x];
    ++deg[y];
  }
  int mnd = deg[1];
  for (i = 1; i <= n1 + n2; ++i) mnd = min(mnd, deg[i]);
  for (i = 1; i <= n1; ++i) adde(S, i, deg[i] - mnd), pos[i] = ed - 1;
  for (i = 1; i <= n2; ++i)
    adde(i + n1, T, deg[i + n1] - mnd), pos[i + n1] = ed - 1;
  for (k = mnd; ~k; --k) {
    if (k != mnd) {
      for (i = 1; i <= n1 + n2; ++i) e[pos[i]].w++;
    }
    for (i = 1; i <= tot; ++i) gap[i] = h[i] = 0;
    for (gap[0] = tot; h[S] < tot; sap(S, inf))
      ;
    for (i = 1; i <= m; ++i)
      if (e[i * 2].w) ans[k].push_back(i);
  }
  for (i = 0; i <= mnd; ++i) {
    printf("%d", ans[i].size());
    for (j = 0; j < ans[i].size(); ++j) printf(" %d", ans[i][j]);
    puts("");
  }
  return 0;
}