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;
int d1[2001], d2[2001];
int ans[2001][2001];
int gae[2001];
struct maxflow {
  maxflow() {}
  struct edge {
    int from;
    int to;
    int flow;
  };
  int n;
  int source, sink;
  vector<int> lvl;
  vector<edge> edges;
  vector<vector<int> > eg;
  queue<int> q;
  vector<int> prv;
  vector<int> sv;
  void init(int _n) {
    n = _n;
    edges.clear();
    lvl.clear();
    lvl.resize(n);
    eg.clear();
    eg.resize(n);
    prv.clear();
    prv.resize(n);
    sv.clear();
    sv.resize(n);
  }
  void add(int from, int to, int flow) {
    int idx = (int)edges.size();
    edges.push_back({from, to, flow});
    edges.push_back({to, from, 0});
    eg[from].push_back(idx);
    eg[to].push_back(idx + 1);
  }
  int bfs() {
    for (int i = 0; i < n; i++) lvl[i] = 1000000000;
    lvl[source] = 0;
    q.push(source);
    while (!q.empty()) {
      int me = q.front();
      q.pop();
      for (const int &i : eg[me]) {
        if ((edges[i]).flow == 0) continue;
        if (lvl[(edges[i]).to] ^ 1000000000) continue;
        lvl[(edges[i]).to] = lvl[me] + 1;
        q.push((edges[i]).to);
      }
    }
    return lvl[sink];
  }
  int backtrack(int x, int lim) {
    if (x == sink) return lim;
    int used = 0;
    for (sv[x]; sv[x] < eg[x].size(); sv[x]++) {
      int i = eg[x][sv[x]];
      if ((edges[i]).flow == 0) continue;
      if (lvl[(edges[i]).to] == lvl[x] + 1) {
        int able = min(lim - used, (edges[i]).flow);
        int res = backtrack((edges[i]).to, able);
        if (res) {
          edges[i].flow -= res;
          edges[i ^ 1].flow += res;
          used += res;
          if (res == able) return res;
        }
      }
    }
    return used;
  }
  int exe(int _source, int _sink) {
    source = _source;
    sink = _sink;
    int ans = 0;
    while (bfs() ^ 1000000000) {
      for (int i = 0; i < n; i++) sv[i] = 0;
      int res = backtrack(source, 1000000000);
      ans += res;
    }
    return ans;
  }
};
int main() {
  int n1, n2, m;
  scanf("%d%d%d", &n1, &n2, &m);
  maxflow mf = maxflow();
  mf.init(n1 + n2 + 2);
  for (int i = 1; i <= n1; i++) {
    mf.add(0, i, 0);
  }
  for (int i = n1 + 1; i <= n1 + n2; i++) {
    mf.add(i, n1 + n2 + 1, 0);
  }
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    mf.add(a, n1 + b, 1);
    d1[a]++;
    d2[b]++;
  }
  int md = d1[1];
  for (int i = 1; i <= n1; i++) md = min(md, d1[i]);
  for (int i = 1; i <= n2; i++) md = min(md, d2[i]);
  for (int i = 1; i <= n1; i++) {
    mf.edges[(i - 1) * 2].flow = d1[i] - md;
  }
  for (int i = 1; i <= n2; i++) {
    mf.edges[2 * n1 + (i - 1) * 2].flow = d2[i] - md;
  }
  for (int t = md; t >= 1; t--) {
    mf.exe(0, n1 + n2 + 1);
    int cntt = 0;
    for (int i = n1 + n2; i < n1 + n2 + m; i++) {
      cntt += (mf.edges[i * 2].flow == 1);
      ans[t][i - (n1 + n2)] = (mf.edges[i * 2].flow == 1);
    }
    for (int i = 1; i <= n1; i++) {
      mf.edges[(i - 1) * 2].flow++;
    }
    for (int i = 1; i <= n2; i++) {
      mf.edges[2 * n1 + (i - 1) * 2].flow++;
    }
    gae[t] = cntt;
  }
  printf("0\n");
  for (int i = 1; i <= md; i++) {
    printf("%d ", gae[i]);
    for (int j = 0; j < m; j++) {
      if (ans[i][j]) printf("%d ", j + 1);
    }
    printf("\n");
  }
}