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 = 4500;
const int M = 6500;
const int inf = 0x3f3f3f3f;
struct Edge {
  int u, v, f, nxt;
} e[M << 1];
int s, t, cnt, head[N], level[N];
queue<int> q;
int n1, n2, m, d[N], id[N], minDegree = inf;
vector<int> ans[M];
void add(int u, int v, int f) {
  e[cnt].u = u;
  e[cnt].v = v;
  e[cnt].f = f;
  e[cnt].nxt = head[u];
  head[u] = cnt++;
  e[cnt].u = v;
  e[cnt].v = u;
  e[cnt].f = 0;
  e[cnt].nxt = head[v];
  head[v] = cnt++;
}
bool bfs() {
  memset(level, -1, sizeof(level));
  level[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 (e[i].f > 0 && level[v] < 0) {
        level[v] = level[u] + 1;
        q.push(v);
      }
    }
  }
  return level[t] > 0;
}
int dfs(int u, int mx) {
  if (u == t || !mx) return mx;
  int tmp, res = 0;
  for (int i = head[u]; i != -1 && mx; i = e[i].nxt) {
    int v = e[i].v;
    if (level[v] == level[u] + 1 && (tmp = dfs(v, min(mx, e[i].f)))) {
      e[i].f -= tmp;
      e[i ^ 1].f += tmp;
      res += tmp;
      mx -= tmp;
    }
  }
  if (!res) level[u] = -1;
  return res;
}
int dinic() {
  int maxflow = 0;
  while (bfs()) {
    maxflow += dfs(s, inf);
  }
  return maxflow;
}
int main() {
  memset(head, -1, sizeof(head));
  cnt = 0;
  scanf("%d%d%d", &n1, &n2, &m);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, n1 + y, 1);
    d[x]++;
    d[n1 + y]++;
    id[i] = cnt - 1;
  }
  for (int i = 1; i <= n1 + n2; i++) minDegree = min(minDegree, d[i]);
  s = 0;
  t = n1 + n2 + 1;
  for (int i = 1; i <= n1; i++) add(s, i, d[i] - minDegree);
  for (int i = 1; i <= n2; i++) add(n1 + i, t, d[n1 + i] - minDegree);
  for (int k = minDegree; k >= 0; k--) {
    dinic();
    for (int i = 1; i <= m; i++)
      if (e[id[i]].f == 0) ans[k].push_back(i);
    for (int i = head[s]; i != -1; i = e[i].nxt) e[i].f++;
    for (int i = head[t]; i != -1; i = e[i].nxt) e[i ^ 1].f++;
  }
  for (int k = 0; k <= minDegree; k++) {
    printf("%d ", ans[k].size());
    for (int i = 0; i < ans[k].size(); i++) printf("%d ", ans[k][i]);
    puts("");
  }
  return 0;
}