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 MAXN_V = 1e4 + 10;
const int MAXN_E = 1e5 + 10;
struct vec {
  int u, next, cap;
  int id;
} edge[MAXN_E];
int num_e;
int head[MAXN_V];
inline void init() {
  for (int i = 0; i < MAXN_V; i++) head[i] = -1;
  num_e = 0;
}
void add(int id, int u, int v, int cap) {
  edge[num_e] = (vec){v, head[u], cap, id};
  head[u] = num_e++;
}
int dep[MAXN_V];
bool dinic_bfs(int s, int t) {
  for (int i = 0; i < MAXN_V; i++) dep[i] = -1;
  queue<int> Q;
  Q.push(s);
  dep[s] = 0;
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (int i = head[u]; ~i; i = edge[i].next) {
      int v = edge[i].u;
      if (dep[v] == -1 && edge[i].cap) {
        dep[v] = dep[u] + 1;
        Q.push(v);
      }
    }
  }
  return ~dep[t];
}
int cur[MAXN_V];
int dinic_dfs(int u, int t, int limit) {
  if (u == t || limit == 0) return limit;
  int f, flow = 0;
  for (int &i = cur[u]; ~i; i = edge[i].next) {
    int v = edge[i].u;
    if (dep[v] == dep[u] + 1 &&
        (f = dinic_dfs(v, t, min(limit, edge[i].cap))) > 0) {
      edge[i].cap -= f;
      edge[i ^ 1].cap += f;
      flow += f;
      if (!(limit -= f)) break;
    }
  }
  return flow;
}
int dinic(int s, int t) {
  int maxflow = 0;
  while (dinic_bfs(s, t)) {
    for (int i = 0; i < MAXN_V; i++) {
      cur[i] = head[i];
    }
    maxflow += dinic_dfs(s, t, INF);
  }
  return maxflow;
}
int n1, n2, m;
int S, T;
int x, y;
int mn;
int deg[MAXN_V];
vector<int> ans[MAXN_V];
int main() {
  scanf("%d%d%d", &n1, &n2, &m);
  S = 0;
  T = n1 + n2 + 1;
  mn = INF;
  for (int i = 0; i <= n1 + n2; i++) deg[i] = 0;
  init();
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &x, &y);
    add(1, x, n1 + y, 1);
    add(1, y + n1, x, 0);
    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++) {
    add(1, S, i, deg[i] - mn);
    add(1, i, S, 0);
  }
  for (int i = 1; i <= n2; i++) {
    add(1, i + n1, T, deg[i + n1] - mn);
    add(1, T, i + n1, 0);
  }
  for (int i = mn; i >= 0; i--) {
    dinic(S, T);
    for (int j = 0; j < m; j++) {
      if (edge[j << 1].cap == 1) ans[i].push_back(j + 1);
    }
    for (int j = head[S]; ~j; j = edge[j].next) {
      edge[j].cap++;
    }
    for (int j = head[T]; ~j; j = edge[j].next) {
      edge[j ^ 1].cap++;
    }
  }
  for (int i = 0; i <= mn; i++) {
    x = ans[i].size();
    printf("%d", x);
    for (int j = 0; j < x; j++) {
      printf(" %d", ans[i][j]);
    }
    printf("\n");
  }
  return 0;
}