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 inf = 0x3f3f3f3f;
const int maxn = 100005;
struct edge {
  int to, re, c, id;
};
vector<edge> g[maxn];
int n, m, s, t;
int n1, n2;
int iter[maxn];
int d[maxn];
int mind = inf;
void add_edge(int x, int y, int cap, int id) {
  g[x].push_back((edge){y, (int)(g[y]).size(), cap, id});
  g[y].push_back((edge){x, (int)(g[x]).size() - 1, 0, -1});
  return;
}
void init() {
  cin >> n1 >> n2 >> m;
  n = n1 + n2 + 2;
  s = 0;
  t = n1 + n2 + 1;
  int x, y;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    add_edge(x, y + n1, 1, i + 1);
  }
  for (int i = 1; i < n1 + n2 + 1; i++) {
    mind = min(mind, (int)(g[i]).size());
  }
  for (int i = 1; i < n1 + 1; i++) {
    add_edge(s, i, (int)(g[i]).size() - mind, -1);
  }
  for (int i = n1 + 1; i < n1 + n2 + 1; i++) {
    add_edge(i, t, (int)(g[i]).size() - mind, -1);
  }
  memset(iter, 0, sizeof(int) * (n + 1));
  memset(d, -1, sizeof(int) * (n + 1));
}
void bfs() {
  queue<int> q;
  while (!q.empty()) q.pop();
  q.push(s);
  d[s] = 0;
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (int i = 0; i < (int)(g[v]).size(); i++) {
      edge &e = g[v][i];
      if (d[e.to] < 0 && e.c > 0) {
        d[e.to] = d[v] + 1;
        q.push(e.to);
      }
    }
  }
}
int dfs(int x, int f) {
  if (x == t || !f) return f;
  int sf = 0;
  for (int &i = iter[x]; i < (int)(g[x]).size(); i++) {
    edge &e = g[x][i];
    if (d[e.to] > d[x] && e.c > 0) {
      int flow = dfs(e.to, min(f, e.c));
      if (flow > 0) {
        e.c -= flow;
        g[e.to][e.re].c += flow;
        sf += flow;
        f -= flow;
        if (!f) break;
      }
    }
  }
  return sf;
}
vector<int> ans[maxn];
void print_path(int k) {
  for (int i = 1; i < n1 + 1; i++) {
    for (int j = 0; j < (int)(g[i]).size(); j++) {
      edge e = g[i][j];
      if (e.id != -1 && e.c) ans[k].push_back(e.id);
    }
  }
}
void max_flow() {
  int flow = 0;
  int k = mind;
  while (k--) {
    for (;;) {
      memset(d, -1, sizeof(int) * (n + 1));
      bfs();
      if (d[t] < 0) break;
      memset(iter, 0, sizeof(int) * (n + 1));
      flow += dfs(s, inf);
    }
    print_path(k + 1);
    for (int i = 0; i < (int)(g[s]).size(); i++) {
      g[s][i].c++;
    }
    for (int i = 0; i < (int)(g[t]).size(); i++) {
      g[g[t][i].to][g[t][i].re].c++;
    }
  }
}
int main() {
  init();
  max_flow();
  for (int i = 0; i < mind + 1; i++) {
    cout << (int)(ans[i]).size() << ' ';
    for (int j = 0; j < (int)(ans[i]).size(); j++) cout << ans[i][j] << ' ';
    cout << endl;
  }
  return 0;
}