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 mod = 1e9 + 7;
const int N = 4010;
int head[N], cnt = 1, dis[N], s, t, cur[N], in[N];
struct edge {
  int next, to, w;
} e[N << 5];
void add(int u, int v, int w) {
  e[++cnt].next = head[u];
  e[cnt].to = v;
  e[cnt].w = w;
  head[u] = cnt;
}
void add_edge(int u, int v, int w) {
  add(u, v, w);
  add(v, u, 0);
}
bool bfs(int s) {
  for (int i = 1; i <= t; i++) dis[i] = -1;
  dis[s] = 0;
  queue<int> q;
  q.push(s);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = head[u]; i; i = e[i].next) {
      int v = e[i].to;
      if (e[i].w && dis[v] == -1) {
        dis[v] = dis[u] + 1;
        q.push(v);
      }
    }
  }
  return dis[t] != -1;
}
int dfs(int u, int flo) {
  if (u == t) return flo;
  int detla = flo;
  for (int i = cur[u]; i; i = e[i].next) {
    cur[u] = e[i].next;
    int v = e[i].to;
    if (e[i].w && dis[v] == dis[u] + 1) {
      int d = dfs(v, min(detla, e[i].w));
      e[i].w -= d;
      e[i ^ 1].w += d;
      detla -= d;
      if (detla == 0) break;
    }
  }
  return flo - detla;
}
int dini() {
  int ans = 0;
  while (bfs(s)) {
    for (int i = 1; i <= t; i++) cur[i] = head[i];
    ans += dfs(s, INF);
  }
  return ans;
}
vector<vector<int>> ans;
int main() {
  int n1, n2, m;
  scanf("%d %d %d", &n1, &n2, &m);
  s = n1 + n2 + 1, t = n1 + n2 + 2;
  for (int i = 1; i <= m; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    b += n1;
    add_edge(a, b, 1);
    in[a]++;
    in[b]++;
  }
  int now = cnt, mx = INF;
  for (int i = 1; i <= n1 + n2; i++) mx = min(mx, in[i]);
  for (int i = 1; i <= n1; i++) add_edge(s, i, in[i] - mx);
  for (int i = n1 + 1; i <= n1 + n2; i++) add_edge(i, t, in[i] - mx);
  for (int i = mx; i >= 0; i--) {
    dini();
    vector<int> res;
    for (int j = 3; j <= now; j += 2) {
      if (e[j].w == 0) {
        res.push_back(j / 2);
      }
    }
    ans.push_back(res);
    for (int i = 1; i <= n1; i++) add_edge(s, i, 1);
    for (int i = n1 + 1; i <= n1 + n2; i++) add_edge(i, t, 1);
  }
  reverse(ans.begin(), ans.end());
  int siz = ans.size();
  for (int i = 0; i < siz; i++) {
    int si = ans[i].size();
    printf("%d", si);
    for (int j = 0; j < si; j++) printf(" %d", ans[i][j]);
    putchar('\n');
  }
  return 0;
}