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 = 2077;
const int INF = 1E9 + 7;
const int SS = maxn * 2 - 1;
const int TT = maxn * 2 - 2;
struct EDGE {
  int to, c, next;
} edges[maxn * 6];
int cEdge = 2, head[maxn * 2];
void addEdge(int from, int to, int c) {
  edges[cEdge] = (EDGE){to, c, head[from]};
  head[from] = cEdge++;
  edges[cEdge] = (EDGE){from, 0, head[to]};
  head[to] = cEdge++;
}
int N1, N2, M, D1[maxn], D2[maxn], E[maxn], U1[maxn], U2[maxn], ca[maxn],
    ans[maxn][maxn];
bool vis[maxn * 2];
bool dfs(int u) {
  if (u == TT) return true;
  vis[u] = true;
  for (int k = head[u]; k; k = edges[k].next) {
    int v = edges[k].to, c = edges[k].c;
    if (!c) continue;
    if (vis[v]) continue;
    if (dfs(v)) {
      edges[k].c--;
      edges[k ^ 1].c++;
      return true;
    }
  }
  return false;
}
int main() {
  scanf("%d%d%d", &N1, &N2, &M);
  for (int i = 1; i <= M; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    ++D1[u];
    ++D2[v];
    E[i] = cEdge;
    addEdge(u, N1 + v, 1);
  }
  int K = INF;
  for (int i = 1; i <= N1; ++i) K = min(K, D1[i]);
  for (int i = 1; i <= N2; ++i) K = min(K, D2[i]);
  for (int i = 1; i <= N1; ++i) {
    U1[i] = cEdge;
    addEdge(SS, i, D1[i] - K);
  }
  for (int i = 1; i <= N2; ++i) {
    U2[i] = cEdge;
    addEdge(N1 + i, TT, D2[i] - K);
  }
  for (int k = K; k >= 0; --k) {
    memset(vis, 0, sizeof(vis));
    while (dfs(SS)) memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= M; ++i)
      if (edges[E[i]].c) ans[k][ca[k]++] = i;
    for (int i = 1; i <= N1; ++i) edges[U1[i]].c++;
    for (int i = 1; i <= N2; ++i) edges[U2[i]].c++;
  }
  for (int k = 0; k <= K; ++k) {
    printf("%d", ca[k]);
    for (int i = 0; i < ca[k]; ++i) printf(" %d", ans[k][i]);
    printf("\n");
  }
}