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 n1, n2, m, lev[4010], cur[4010], h[4010], num = 1, T = 4005, du[4010],
                                              K = 0x3f3f3f3f;
vector<int> ans[4010 >> 1];
struct edge {
  int to, next, val;
} data[12100];
inline void add(int x, int y, int val) {
  data[++num].to = y;
  data[num].next = h[x];
  h[x] = num;
  data[num].val = val;
  data[++num].to = x;
  data[num].next = h[y];
  h[y] = num;
  data[num].val = 0;
}
inline bool bfs() {
  queue<int> q;
  memset(lev, 0, sizeof(lev));
  q.push(0);
  lev[0] = 1;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = h[x]; i; i = data[i].next) {
      int y = data[i].to;
      if (lev[y] || !data[i].val) continue;
      lev[y] = lev[x] + 1;
      if (y == T) return 1;
      q.push(y);
    }
  }
  return 0;
}
inline int dinic(int x, int low) {
  if (x == T) return low;
  int tmp = low;
  for (int &i = cur[x]; i; i = data[i].next) {
    int y = data[i].to;
    if (lev[y] != lev[x] + 1 || !data[i].val) continue;
    int res = dinic(y, min(tmp, data[i].val));
    if (!res)
      lev[y] = 0;
    else
      tmp -= res, data[i].val -= res, data[i ^ 1].val += res;
    if (!tmp) return low;
  }
  return low - tmp;
}
int main() {
  cin >> n1 >> n2 >> m;
  for (int i = 1; i <= m; ++i) {
    int x, y;
    cin >> x >> y;
    add(x, y + n1, 1);
    du[x]++;
    du[y + n1]++;
  }
  for (int i = 1; i <= n1 + n2; ++i) K = min(K, du[i]);
  for (int i = 1; i <= n1; ++i)
    if (du[i] > K) add(0, i, du[i] - K);
  for (int i = 1; i <= n2; ++i)
    if (du[i + n1] > K) add(i + n1, T, du[i + n1] - K);
  for (int j = K; j >= 1; --j) {
    while (bfs()) {
      memcpy(cur, h, sizeof(h));
      dinic(0, 0x3f3f3f3f);
    }
    for (int i = 1; i <= m; ++i)
      if (data[i << 1].val) ans[j].push_back(i);
    for (int i = 1; i <= n1; ++i) add(0, i, 1);
    for (int i = 1; i <= n2; ++i) add(i + n1, T, 1);
  }
  puts("0");
  for (int i = 1; i <= K; ++i) {
    printf("%d", ans[i].size());
    for (int j = 0; j < ans[i].size(); ++j) printf(" %d", ans[i][j]);
    puts("");
  }
  return 0;
}