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 = 4e3 + 10;
const int inf = 1e8;
struct edge {
  int to, cap, next;
} e[N * 10];
int n1, n2, m, mindeg, cnt, head[N], level[N], iter[N], deg[N];
inline void add(int from, int to, int cap) {
  e[++cnt].to = to, e[cnt].cap = cap, e[cnt].next = head[from],
  head[from] = cnt;
  e[++cnt].to = from, e[cnt].cap = 0, e[cnt].next = head[to], head[to] = cnt;
}
int dfs(int v, int t, int f) {
  if (v == t) return f;
  for (int &i = iter[v]; i; i = e[i].next) {
    int u = e[i].to;
    if (e[i].cap > 0 && level[u] > level[v]) {
      int d = dfs(u, t, min(e[i].cap, f));
      if (d > 0) {
        e[i].cap -= d;
        e[i ^ 1].cap += d;
        return d;
      }
    }
  }
  return 0;
}
bool bfs(int s, int t) {
  memset(level, -1, sizeof(level));
  level[s] = 0;
  queue<int> que;
  que.push(s);
  while (!que.empty()) {
    int v = que.front();
    que.pop();
    for (int i = head[v]; i; i = e[i].next) {
      if (e[i].cap > 0 && level[e[i].to] < 0) {
        level[e[i].to] = level[v] + 1;
        que.push(e[i].to);
      }
    }
  }
  return level[t] >= 0;
}
int dinic(int s, int t) {
  int flow = 0;
  while (bfs(s, t)) {
    for (int i = 1; i < N; i++) iter[i] = head[i];
    int f;
    while ((f = dfs(s, t, inf)) > 0) {
      flow += f;
    }
  }
  return flow;
}
int main() {
  scanf("%d%d%d", &n1, &n2, &m);
  cnt = 1, mindeg = inf;
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    deg[u]++, deg[v + n1]++;
    add(u, v + n1, 1);
  }
  for (int i = 1; i <= n1 + n2; i++) {
    mindeg = min(mindeg, deg[i]);
  }
  int s = n1 + n2 + 1, t = s + 1;
  for (int i = 1; i <= n1; i++) add(s, i, 0);
  for (int i = 1; i <= n2; i++) add(i + n1, t, 0);
  for (int k = 0; k <= mindeg; k++) {
    for (int i = 2 * m + 2; i < cnt; i += 2) {
      int x = e[i ^ 1].to;
      if (x > n1 + n2) x = e[i].to;
      e[i].cap = deg[x] - k, e[i ^ 1].cap = 0;
    }
    for (int i = 2; i <= 2 * m; i += 2) {
      e[i].cap = 1, e[i ^ 1].cap = 0;
    }
    printf("%d ", m - dinic(s, t));
    for (int x = 1; x <= n1; x++) {
      for (int i = head[x]; i; i = e[i].next) {
        if (e[i].to != s && e[i].cap) {
          printf("%d ", i / 2);
        }
      }
    }
    printf("\n");
  }
  return 0;
}