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 MAXN = 2010;
struct Edge {
  int ver, nxt, flow;
} e[MAXN * 20];
int head[MAXN * 2], tot = 1, S = 0, T = 4001;
void add(int u, int v, int f) {
  e[++tot].ver = v;
  e[tot].flow = f;
  e[tot].nxt = head[u];
  head[u] = tot;
}
queue<int> q;
int dep[MAXN * 2], d[MAXN * 2];
bool bfs() {
  memset(dep, -1, sizeof(dep));
  q.push(S);
  dep[S] = 0;
  int i;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (i = head[u]; i; i = e[i].nxt) {
      int v = e[i].ver, f = e[i].flow;
      if (!(~dep[v]) && f) {
        dep[v] = dep[u] + 1;
        q.push(v);
      }
    }
  }
  return ~dep[T];
}
int dfs(int u, int f) {
  int res = 0, i;
  if (u == T) return f;
  for (i = head[u]; i; i = e[i].nxt) {
    int v = e[i].ver, fl = e[i].flow;
    if (dep[v] == dep[u] + 1 && fl) {
      int F = dfs(v, min(f - res, fl));
      e[i].flow -= F;
      e[i ^ 1].flow += F;
      res += F;
      if (res == f) return f;
    }
  }
  if (res == 0) dep[u] = -1;
  return res;
}
int Dinic() {
  int res = 0;
  while (bfs()) res += dfs(S, INT_MAX);
  return res;
}
int num[MAXN], u[MAXN], v[MAXN];
vector<int> res[MAXN];
int main() {
  int n1, n2, m, minn = INT_MAX, i, j;
  scanf("%d %d %d", &n1, &n2, &m);
  for (i = 1; i <= m; i++) {
    scanf("%d %d", &u[i], &v[i]);
    v[i] += n1;
    d[u[i]]++;
    d[v[i]]++;
  }
  for (i = 1; i <= n1 + n2; i++) minn = min(minn, d[i]);
  for (i = 1; i <= n1; i++) {
    add(S, i, d[i] - minn);
    add(i, S, 0);
  }
  for (i = n1 + 1; i <= n1 + n2; i++) {
    add(i, T, d[i] - minn);
    add(T, i, 0);
  }
  for (i = 1; i <= m; i++) {
    add(u[i], v[i], 1);
    add(v[i], u[i], 0);
    num[i] = tot;
  }
  for (i = minn; i >= 0; i--) {
    Dinic();
    for (j = 1; j <= m; j++) {
      if (!e[num[j]].flow) res[i].push_back(j);
    }
    for (j = head[S]; j; j = e[j].nxt) e[j].flow++;
    for (j = head[T]; j; j = e[j].nxt) e[j ^ 1].flow++;
  }
  for (i = 0; i <= minn; i++) {
    printf("%d ", res[i].size());
    for (j = res[i].size() - 1; j >= 0; j--) printf("%d ", res[i][j]);
    printf("\n");
  }
  return 0;
}