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 = 4e3 + 5, inf = 1e9;
vector<int> ans[N];
int n1, n2, mn = inf, m, s, t;
struct yal {
  int to, flow, C, ind, rev;
};
vector<yal> adj[N];
struct FreeRun {
  int CNT[N], H[N], q[N], l, r;
  void add(int u, int v, int C, int ind) {
    yal a{v, 0, C, ind, (int)adj[v].size()};
    yal b{u, 0, 0, -ind, (int)adj[u].size()};
    adj[u].push_back(a);
    adj[v].push_back(b);
  }
  bool cork(int s, int t) {
    memset(H, -1, sizeof(H));
    H[s] = l = r = 0;
    q[r++] = s;
    while (l ^ r) {
      int u = q[l++];
      for (yal &y : adj[u]) {
        if (H[y.to] < 0 && y.flow < y.C) {
          H[y.to] = H[u] + 1;
          q[r++] = y.to;
        }
      }
    }
    return (H[t] < 0 ? 0 : 1);
  }
  int wallspin(int flow, int u, int t) {
    if (u == t) return (flow);
    for (; CNT[u] < (int)adj[u].size(); CNT[u]++) {
      yal &y = adj[u][CNT[u]];
      if (!(H[y.to] == H[u] + 1 && y.flow < y.C)) continue;
      int tmp = min(flow, y.C - y.flow);
      int ah = wallspin(tmp, y.to, t);
      if (ah <= 0) continue;
      y.flow += ah;
      adj[y.to][y.rev].flow -= ah;
      return (ah);
    }
    return (0);
  }
  inline int gainer(int s, int t) {
    if (s == t) return (-1);
    int rt = 0;
    while (cork(s, t)) {
      memset(CNT, 0, sizeof(CNT));
      while (int tmp = wallspin(inf, s, t)) rt += tmp;
    }
    return (rt);
  }
} f1;
int main() {
  scanf("%d %d %d", &n1, &n2, &m);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    f1.add(a, n1 + b, 1, i + 1);
  }
  t = n1 + n2 + 1;
  for (int i = 1; i < t; i++) mn = min(mn, (int)adj[i].size());
  for (int i = 1; i <= n1; i++) f1.add(s, i, (int)adj[i].size() - mn, -1);
  for (int i = n1 + 1; i < t; i++) f1.add(i, t, (int)adj[i].size() - mn, -1);
  for (int i = mn; ~i; i--) {
    f1.gainer(s, t);
    for (int j = 1; j <= n1; j++)
      for (yal y : adj[j])
        if (y.ind > 0 && y.to && !y.flow) ans[i].push_back(y.ind);
    for (int j = 1; j <= n1; j++) f1.add(s, j, 1, -1);
    for (int j = n1 + 1; j < t; j++) f1.add(j, t, 1, -1);
  }
  for (int i = 0; i <= mn; i++) {
    printf("%d ", (int)ans[i].size());
    for (int x : ans[i]) printf("%d ", x);
    puts("");
  }
}