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;
template <class t>
inline t read(t &x) {
  char c = getchar();
  bool f = 0;
  x = 0;
  while (!isdigit(c)) f |= c == '-', c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  if (f) x = -x;
  return x;
}
template <class t>
inline void write(t x) {
  if (x < 0)
    putchar('-'), write(-x);
  else {
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
  }
}
const int N = 4005, M = 2005;
int n1, n2, m, k, du[N], use[N], h[N];
bool v[N], fl;
vector<int> ans;
struct edge {
  int n, v, f;
} e[M << 1];
void add(int i, int x, int y, int f) {
  e[i] = {h[x], y, f};
  h[x] = i;
}
void dfs(int x) {
  v[x] = 1;
  if (x > n1 && use[x] < du[x] - k) {
    fl = 1;
    use[x]++;
    return;
  }
  for (int i = h[x]; i; i = e[i].n) {
    int y = e[i].v;
    if (v[y] || e[i].f == 0) continue;
    e[i].f = 0;
    e[i ^ 1].f = 1;
    dfs(y);
    if (fl) break;
    e[i].f = 1;
    e[i ^ 1].f = 0;
  }
}
void flow(int x) {
  memset(v, 0, sizeof v);
  fl = 0;
  dfs(x);
}
signed main() {
  read(n1);
  read(n2);
  read(m);
  bool fl = 0;
  if (n1 < n2) fl = 1, swap(n1, n2);
  for (int i = 1, x, y; i <= m; i++) {
    read(x), read(y);
    if (fl) swap(x, y);
    y += n1;
    du[x]++;
    du[y]++;
    add(i << 1, x, y, 1);
    add(i << 1 | 1, y, x, 0);
  }
  int D = *min_element(du + 1, du + 1 + n1 + n2);
  for (k = D; k; k--) {
    for (int i = 1; i <= n1; i++) {
      if (k == D)
        for (int j = 1; j <= du[i] - k; j++) flow(i);
      else
        flow(i);
    }
    for (int i = 1; i <= m; i++)
      if (e[i << 1].f) ans.push_back(i);
  }
  for (int i = 0, t; i <= D; i++) {
    write(t = i * max(n1, n2));
    putchar(' ');
    while (t--) write(ans.back()), ans.pop_back(), putchar(' ');
    puts("");
  }
}