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 = 1145e4;
struct qwq {
  int v;
  int flow;
  int nxt;
} edge[12010];
int cnt = -1;
int head[12010];
void add(int u, int v, int w) {
  edge[++cnt].nxt = head[u];
  edge[cnt].v = v;
  edge[cnt].flow = w;
  head[u] = cnt;
}
void addedge(int u, int v, int w) {
  add(u, v, w);
  add(v, u, 0);
}
int dep[12010];
int cur[12010];
bool bfs(int s, int t) {
  memset(dep, 0, sizeof(dep));
  queue<int> q;
  q.push(s);
  dep[s] = 1;
  cur[s] = head[s];
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = head[u]; ~i; i = edge[i].nxt) {
      int v = edge[i].v;
      if (edge[i].flow > 0 && !dep[v]) {
        dep[v] = dep[u] + 1;
        cur[v] = head[v];
        q.push(v);
      }
    }
  }
  return dep[t];
}
int dfs(int u, int t, int lim) {
  if (u == t || lim == 0) return lim;
  int flow = 0, tmp;
  for (int i = cur[u]; ~i; i = edge[i].nxt) {
    cur[u] = i;
    int v = edge[i].v, w = edge[i].flow;
    if (dep[v] == dep[u] + 1 && (tmp = dfs(v, t, min(w, lim - flow)))) {
      flow += tmp;
      edge[i].flow -= tmp;
      edge[i ^ 1].flow += tmp;
      if (!(lim - flow)) break;
    }
  }
  return flow;
}
int ret;
int dinic(int s, int t) {
  while (bfs(s, t)) {
    ret += dfs(s, t, inf);
  }
  return ret;
}
int S, T;
int n1, n2, m;
int dg[12010];
int k;
int ans[12010];
vector<int> v[12010];
int mm;
int main() {
  memset(head, -1, sizeof(head));
  scanf("%d%d%d", &n1, &n2, &m);
  S = 0, T = n1 + n2 + 1;
  for (int i = 1; i <= m; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    dg[u]++, dg[v + n1]++;
    addedge(u, v + n1, 1);
  }
  mm = cnt;
  k = inf;
  for (int i = 1; i <= n1 + n2; ++i) {
    k = min(k, dg[i]);
  }
  int kk = k;
  for (int i = 1; i <= n1; ++i) {
    addedge(S, i, dg[i] - k);
  }
  for (int i = 1; i <= n2; ++i) {
    addedge(i + n1, T, dg[n1 + i] - k);
  }
  while (k >= 0) {
    ans[k] = m - dinic(S, T);
    for (int i = 0; i <= mm; i += 2) {
      if (edge[i].flow) {
        v[k].push_back(i / 2 + 1);
      }
    }
    for (int i = head[S]; ~i; i = edge[i].nxt) {
      edge[i].flow++;
    }
    for (int i = head[T]; ~i; i = edge[i].nxt) {
      edge[i ^ 1].flow++;
    }
    k--;
  }
  for (int i = 0; i <= kk; ++i) {
    printf("%d ", ans[i]);
    for (int j = 0; j < ans[i]; ++j) {
      printf("%d ", v[i][j]);
    }
    puts("");
  }
}