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 DIM = 4005;
const int INF = 0x3f3f3f3f;
int cap[DIM][DIM], fth[DIM], dis[DIM], deg[DIM];
deque<int> que;
vector<int> edg[DIM], edl[DIM / 2][DIM / 2], ans[DIM];
bool bfs(int src, int dst) {
  memset(dis, INF, sizeof dis);
  que.assign(1, src);
  dis[src] = 1;
  for (; !que.empty(); que.pop_front()) {
    int x = que.front();
    for (int y : edg[x])
      if (dis[y] > dis[x] + 1 and cap[x][y] > 0) {
        dis[y] = dis[x] + 1;
        fth[y] = x;
        if (y != dst) que.push_back(y);
      }
  }
  return dis[dst] != INF;
}
int main(void) {
  int n1, n2, m;
  cin >> n1 >> n2 >> m;
  for (int i = 1; i <= m; ++i) {
    int x, y;
    cin >> x >> y;
    y += n1;
    if (edl[x][y - n1].empty()) {
      edg[x].push_back(y);
      edg[y].push_back(x);
    }
    ++cap[x][y];
    edl[x][y - n1].push_back(i);
  }
  int src = n1 + n2 + 1, dst = n1 + n2 + 2, mnd = INF;
  for (int i = 1; i <= n1; ++i) {
    for (int j = n1 + 1; j <= n1 + n2; ++j) deg[i] += cap[i][j];
    mnd = min(mnd, deg[i]);
  }
  for (int i = n1 + 1; i <= n1 + n2; ++i) {
    for (int j = 1; j <= n1; ++j) deg[i] += cap[j][i];
    mnd = min(mnd, deg[i]);
  }
  for (int i = 1; i <= n1 + n2; ++i) {
    if (i <= n1) {
      cap[src][i] = deg[i] - mnd;
      edg[src].push_back(i);
      edg[i].push_back(src);
    } else {
      cap[i][dst] = deg[i] - mnd;
      edg[i].push_back(dst);
      edg[dst].push_back(i);
    }
  }
  for (int k = mnd; k >= 0; --k) {
    while (bfs(src, dst)) {
      for (int x : edg[dst]) {
        if (cap[x][dst] > 0 and dis[x] != INF) {
          fth[dst] = x;
          int mnm = INF;
          for (int y = dst; y != src; y = fth[y])
            mnm = min(mnm, cap[fth[y]][y]);
          for (int y = dst; y != src; y = fth[y]) {
            cap[fth[y]][y] -= mnm;
            cap[y][fth[y]] += mnm;
          }
        }
      }
    }
    for (int i = 1; i <= n1; ++i)
      for (int j = n1 + 1; j <= n1 + n2; ++j)
        for (int p = 0; p < cap[i][j]; ++p) ans[k].push_back(edl[i][j - n1][p]);
    for (int i = 1; i <= n1 + n2; ++i) i <= n1 ? ++cap[src][i] : ++cap[i][dst];
  }
  for (int i = 0; i <= mnd; ++i) {
    cout << ans[i].size() << " ";
    for (int x : ans[i]) cout << x << " ";
    cout << endl;
  }
  return 0;
}