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 maxn = 2020;
const long long INF = 100000000000000LL;
const long long mod = 1e9 + 9;
int n1, n2, m, n;
vector<int> g[maxn * 3];
vector<int> id[maxn * 3];
int flow[maxn * 2][maxn * 2];
int t[maxn * 2][maxn * 2];
int deg[maxn * 2];
int vis[maxn * 2];
vector<int> ans[maxn];
int cnt = 0;
int dfs(int u) {
  if (u == n) return 1;
  vis[u] = cnt;
  for (int i = 0, v; i < g[u].size(); i++) {
    v = g[u][i];
    if (!flow[u][v] || vis[v] == cnt) continue;
    if (dfs(v)) {
      flow[u][v]--;
      flow[v][u]++;
      return 1;
    }
  }
  return 0;
}
void solve() {
  memset(t, 0, sizeof(t));
  memset(deg, 0, sizeof(deg));
  memset(flow, 0, sizeof(flow));
  cin >> n1 >> n2 >> m;
  n = n1 + n2 + 1;
  int mindeg = maxn;
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    v = n1 + v;
    flow[u][v]++;
    deg[u]++, deg[v]++;
    g[u].push_back(v);
    g[v].push_back(u);
    id[u].push_back(i);
    id[v].push_back(i);
  }
  for (int i = 1; i < n; i++) mindeg = min(mindeg, deg[i]);
  for (int i = 1; i <= n1; i++) {
    g[0].push_back(i);
    g[i].push_back(0);
    flow[0][i] = deg[i] - mindeg;
  }
  for (int i = n1 + 1; i < n; i++) {
    g[i].push_back(n);
    g[n].push_back(i);
    flow[i][n] = deg[i] - mindeg;
  }
  for (int kk = mindeg; kk >= 0; kk--) {
    cnt++;
    while (dfs(0)) cnt++;
    for (int i = 1; i <= n1; i++) {
      for (int j = 0; j < g[i].size(); j++) {
        int v = g[i][j];
        t[i][v] = flow[i][v];
      }
    }
    for (int i = 1; i <= n1; i++) {
      for (int j = 0; j < g[i].size(); j++) {
        int v = g[i][j];
        if (v && t[i][v]) {
          ans[kk].push_back(id[i][j]);
          t[i][v]--;
        }
      }
    }
    for (int j = 1; j <= n1; j++) flow[0][j]++;
    for (int j = n1 + 1; j < n; j++) flow[j][n]++;
  }
  for (int i = 0; i <= mindeg; i++) {
    int cnt = ans[i].size();
    printf("%d", cnt);
    for (int j = 0; j < cnt; j++) printf(" %d", ans[i][j]);
    puts("");
  }
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  int mark_EOF;
  if (scanf("%d", &mark_EOF) == EOF) return 0;
  return 0;
}