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;
struct data {
  int v, w, nxt;
} e[2000010];
int head[200010];
int cnt;
void add(int u, int v, int w) {
  e[++cnt].v = v;
  e[cnt].w = w;
  e[cnt].nxt = head[u];
  head[u] = cnt;
}
inline void ad(int u, int v, int w) {
  add(u, v, w);
  add(v, u, 0);
}
int n1, n2, m, k = 2e9;
int dis[200010];
int cur[200010];
queue<int> q;
int s, t;
void init() { cnt = -1, memset(head, -1, sizeof(head)); }
int bfs() {
  for (int i = s; i <= t; i++) dis[i] = -1, cur[i] = head[i];
  while (!q.empty()) q.pop();
  dis[s] = 0;
  q.push(s);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = head[u]; i != -1; i = e[i].nxt) {
      int v = e[i].v;
      if (dis[v] == -1 && e[i].w) {
        dis[v] = dis[u] + 1;
        q.push(v);
      }
    }
  }
  return dis[t] != -1;
}
int dfs(int u, int exp) {
  if (u == t) return exp;
  int flow = 0, tmp = 0;
  for (int i = cur[u]; i != -1; i = e[i].nxt) {
    cur[u] = i;
    int v = e[i].v;
    if (dis[v] == dis[u] + 1 && e[i].w) {
      tmp = dfs(v, min(exp, e[i].w));
      if (!tmp) continue;
      exp -= tmp;
      flow += tmp;
      e[i].w -= tmp;
      e[i ^ 1].w += tmp;
      if (!exp) break;
    }
  }
  return flow;
}
void calc() {
  while (bfs()) dfs(s, 0x7fffffff);
}
vector<int> ans[2010];
int id[100010];
int d[100010];
int main() {
  init();
  int x, y;
  scanf("%d%d%d", &n1, &n2, &m);
  s = 0, t = n1 + n2 + 1;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &x, &y);
    ad(x, n1 + y, 1);
    id[i] = cnt;
    d[x]++, d[n1 + y]++;
  }
  for (int i = 1; i <= n1 + n2; i++) k = min(k, d[i]);
  for (int i = 1; i <= n1; i++) ad(s, i, d[i] - k);
  for (int i = n1 + 1; i <= n1 + n2; i++) ad(i, t, d[i] - k);
  for (int i = k; i >= 0; i--) {
    calc();
    for (int j = 1; j <= m; j++)
      if (e[id[j]].w == 0) ans[i].push_back(j);
    for (int j = head[s]; j != -1; j = e[j].nxt) e[j].w++;
    for (int j = head[t]; j != -1; j = e[j].nxt) e[j ^ 1].w++;
  }
  for (int i = 0; i <= k; i++) {
    printf("%d ", ans[i].size());
    for (int j = 0; j < ans[i].size(); j++) printf("%d ", ans[i][j]);
    printf("\n");
  }
}