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, n2, m, deg[5010], id[5010], mn = 0x3f3f3f3f;
namespace MaxFlow {
const int N = 5000, M = 2000000;
int head[N], cur[N], dep[N], cnt, S, T, res;
struct node {
  int to, next, val;
} edge[M];
void ae(int u, int v, int w) {
  edge[cnt].next = head[u], edge[cnt].to = v, edge[cnt].val = w,
  head[u] = cnt++;
  edge[cnt].next = head[v], edge[cnt].to = u, edge[cnt].val = 0,
  head[v] = cnt++;
}
queue<int> q;
inline bool bfs() {
  memset(dep, 0, sizeof(dep)), q.push(S), dep[S] = 1;
  while (!q.empty()) {
    register int x = q.front();
    q.pop();
    for (register int i = cur[x] = head[x]; i != -1; i = edge[i].next)
      if (edge[i].val && !dep[edge[i].to])
        dep[edge[i].to] = dep[x] + 1, q.push(edge[i].to);
  }
  return dep[T] > 0;
}
bool reach;
inline int dfs(int x, int flow) {
  if (x == T) {
    res += flow;
    reach = true;
    return flow;
  }
  int used = 0;
  for (register int &i = cur[x]; i != -1; i = edge[i].next) {
    if (!edge[i].val || dep[edge[i].to] != dep[x] + 1) continue;
    register int ff = dfs(edge[i].to, min(edge[i].val, flow - used));
    if (ff) {
      edge[i].val -= ff;
      edge[i ^ 1].val += ff;
      used += ff;
      if (used == flow) break;
    }
  }
  return used;
}
inline void Dinic() {
  while (bfs()) {
    reach = true;
    while (reach) reach = false, dfs(S, 0x3f3f3f3f);
  }
}
}  // namespace MaxFlow
using namespace MaxFlow;
vector<int> v[5010];
int main() {
  scanf("%d%d%d", &n1, &n2, &m), memset(head, -1, sizeof(head)),
      S = n1 + n2 + 1, T = n1 + n2 + 2;
  for (int i = 1, x, y; i <= m; i++)
    scanf("%d%d", &x, &y), ae(x, y + n1, 1), deg[x]++, deg[y + n1]++;
  for (int i = 1; i <= n1 + n2; i++) mn = min(mn, deg[i]);
  for (int i = 1; i <= n1; i++) id[i] = cnt, ae(S, i, deg[i] - mn);
  for (int i = n1 + 1; i <= n1 + n2; i++) id[i] = cnt, ae(i, T, deg[i] - mn);
  for (int i = 0; i <= mn; i++) {
    Dinic();
    for (int j = 0; j < m; j++)
      if (edge[j << 1].val) v[i].push_back(j + 1);
    for (int j = 1; j <= n1 + n2; j++) edge[id[j]].val++;
  }
  for (int i = mn; i >= 0; i--) {
    printf("%d ", v[i].size());
    for (int j = 0; j < v[i].size(); j++) printf("%d ", v[i][j]);
    puts("");
  }
  return 0;
}