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 n1;
int n2;
int m;
int base1;
int base2;
int star;
int en;
map<int, int> mp;
vector<int> go[5002];
vector<int> rev[5002];
vector<int> cap[5002];
vector<int> id[5002];
void add(int a, int b, int c) {
  go[a].push_back(b);
  cap[a].push_back(c);
  rev[a].push_back(rev[b].size());
  go[b].push_back(a);
  cap[b].push_back(0);
  rev[b].push_back(rev[a].size() - 1);
  id[a].push_back(-1);
  id[b].push_back(-1);
}
bool use[5002];
inline bool dfs(int cur) {
  if (cur == en) {
    return true;
  }
  use[cur] = true;
  for (int i = 0; i < go[cur].size(); i++) {
    if (use[go[cur][i]]) continue;
    if (cap[cur][i] == 0) continue;
    if (dfs(go[cur][i]) == false) continue;
    cap[go[cur][i]][rev[cur][i]]++;
    cap[cur][i]--;
    return true;
  }
  return false;
}
void done() {
  while (1) {
    memset(use, false, sizeof(use));
    bool ok = dfs(star);
    if (ok) {
      continue;
    }
    break;
  }
}
map<pair<int, int>, int> ed;
vector<int> ans[5002];
int main() {
  cin >> n1 >> n2 >> m;
  star = 0;
  en = 1;
  base1 = 2;
  base2 = base1 + n1;
  for (int i = 0; i < n1; i++) {
    mp[base1 + i] = 0;
  }
  for (int i = 0; i < n2; i++) {
    mp[base2 + i] = 0;
  }
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    a += base1;
    b += base2;
    mp[a]++;
    mp[b]++;
    add(a, b, 1);
    ed[make_pair(a - base1, b - base2)] = i + 1;
    id[a].back() = i + 1;
  }
  int mint = INT_MAX;
  for (auto it : mp) {
    mint = min(mint, it.second);
  }
  for (int i = 0; i < n1; i++) {
    add(star, base1 + i, mp[base1 + i] - mint);
  }
  for (int i = 0; i < n2; i++) {
    add(base2 + i, en, mp[base2 + i] - mint);
  }
  bool flag = true;
  int C = mint;
  int cur = -1;
  while (mint--) {
    cur++;
    if (flag == false) {
      for (int i = 0; i < go[star].size(); i++) {
        cap[star][i]++;
      }
      for (int i = 0; i < go[en].size(); i++) {
        cap[go[en][i]][rev[en][i]]++;
      }
    }
    done();
    flag = false;
    for (int i = 0; i < n1; i++) {
      int node = base1 + i;
      for (int j = 0; j < go[node].size(); j++) {
        if (cap[node][j] == 1 && go[node][j] != star) {
          ans[C - cur].push_back(id[node][j]);
        }
      }
    }
  }
  for (int i = 0; i <= C; i++) {
    cout << ans[i].size();
    for (int el : ans[i]) {
      printf(" %d", el);
    }
    puts("");
  }
  return 0;
}