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 = 4010;
vector<int> g[N], num[N], ans[N];
int f[N][N], tem[N][N], n, m, d[N], cnt, v[N];
bool dfs(int x) {
  if (x == n) return true;
  v[x] = cnt;
  for (int i = 0; i < g[x].size(); i++) {
    int y = g[x][i];
    if (f[x][y] == 0 || v[y] == cnt) continue;
    if (dfs(y)) {
      f[x][y]--;
      f[y][x]++;
      return true;
    }
  }
  return false;
}
int main() {
  int n1, n2, x, y, k;
  scanf("%d%d%d", &n1, &n2, &k);
  n = n1 + n2 + 1;
  for (int i = 1; i <= k; i++) {
    scanf("%d%d", &x, &y);
    y += n1;
    g[x].push_back(y);
    g[y].push_back(x);
    num[x].push_back(i);
    f[x][y]++;
    d[x]++, d[y]++;
  }
  m = k * 2;
  for (int i = 1; i < n; i++) m = min(m, d[i]);
  for (int i = 1; i <= n1; i++) {
    g[0].push_back(i);
    g[i].push_back(0);
    f[0][i] = d[i] - m;
  }
  for (int i = n1 + 1; i < n; i++) {
    g[i].push_back(n);
    g[n].push_back(i);
    f[i][n] = d[i] - m;
  }
  for (int l = m; l >= 0; l--) {
    cnt++;
    while (dfs(0)) cnt++;
    for (int i = 1; i <= n1; i++) {
      for (int j = 0; j < g[i].size(); j++) {
        y = g[i][j];
        tem[i][y] = f[i][y];
      }
    }
    for (int i = 1; i <= n1; i++) {
      for (int j = 0; j < g[i].size(); j++) {
        y = g[i][j];
        if (y && tem[i][y]) {
          tem[i][y]--;
          ans[l].push_back(num[i][j]);
        }
      }
    }
    for (int i = 1; i <= n1; i++) f[0][i]++;
    for (int i = n1 + 1; i < n; i++) f[i][n]++;
  }
  for (int i = 0; i <= m; i++) {
    printf("%d", ans[i].size());
    for (auto &x : ans[i]) printf(" %d", x);
    printf("\n");
  }
}