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 = 4050;
const int MAXM = 10005;
const int INF = 1000000050;
int Head[MAXN], cur[MAXN], lev[MAXN], to[MAXM << 1], nxt[MAXM << 1],
    f[MAXM << 1], ed = 1, S, T;
inline void addedge(int u, int v, int cap) {
  to[++ed] = v;
  nxt[ed] = Head[u];
  Head[u] = ed;
  f[ed] = cap;
  to[++ed] = u;
  nxt[ed] = Head[v];
  Head[v] = ed;
  f[ed] = 0;
  return;
}
inline bool BFS() {
  int u;
  memset(lev, -1, sizeof(lev));
  queue<int> q;
  lev[S] = 0;
  q.push(S);
  while (q.size()) {
    u = q.front();
    q.pop();
    for (int i = Head[u]; i; i = nxt[i])
      if (f[i] && lev[to[i]] == -1) {
        lev[to[i]] = lev[u] + 1;
        q.push(to[i]);
      }
  }
  memcpy(cur, Head, sizeof Head);
  return lev[T] != -1;
}
inline int DFS(int u, int maxf) {
  if (u == T || !maxf) {
    return maxf;
  }
  int cnt = 0;
  for (int &i = cur[u], tem; i; i = nxt[i])
    if (f[i] && lev[to[i]] == lev[u] + 1) {
      tem = DFS(to[i], min(maxf, f[i]));
      maxf -= tem;
      f[i] -= tem;
      f[i ^ 1] += tem;
      cnt += tem;
      if (!maxf) {
        break;
      }
    }
  if (!cnt) {
    lev[u] = -1;
  }
  return cnt;
}
int Dinic() {
  int ans = 0;
  while (BFS()) {
    ans += DFS(S, 2147483647);
  }
  return ans;
}
void init(int SS, int TT) {
  memset(Head, 0, sizeof(Head));
  ed = 1;
  S = SS;
  T = TT;
  return;
}
int du[4005];
int ans[4005][4005];
int main() {
  int n1, n2, m;
  int u, v;
  scanf("%d %d %d", &n1, &n2, &m);
  int n = n1 + n2;
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", &u, &v);
    addedge(u, v + n1, 1);
    du[u]++, du[v + n1]++;
  }
  int Mindegree = INT_MAX;
  for (int i = 1; i <= n; i++) {
    Mindegree = min(Mindegree, du[i]);
  }
  S = 0, T = n + 1;
  for (int i = 1; i <= n1; i++) {
    addedge(S, i, du[i] - Mindegree);
  }
  for (int i = n1 + 1; i <= n; i++) {
    addedge(i, T, du[i] - Mindegree);
  }
  int ansnow = Dinic();
  for (int x = 1; x <= n1; x++) {
    for (int i = Head[x]; i; i = nxt[i]) {
      v = to[i];
      if (v >= n1 + 1 && v <= n) {
        if (f[i] == 1) {
          ans[Mindegree][++ans[Mindegree][0]] = i / 2;
        }
      }
    }
  }
  for (int i = Head[S]; i; i = nxt[i]) {
    f[i]++;
  }
  for (int x = n1 + 1; x <= n; x++) {
    for (int i = Head[x]; i; i = nxt[i]) {
      v = to[i];
      if (v == T) {
        f[i]++;
      }
    }
  }
  for (int i = Mindegree - 1; i >= 0; i--) {
    ansnow = Dinic();
    for (int x = 1; x <= n1; x++) {
      for (int j = Head[x]; j; j = nxt[j]) {
        v = to[j];
        if (v >= n1 + 1 && v <= n) {
          if (f[j] == 1) {
            ans[i][++ans[i][0]] = j / 2;
          }
        }
      }
    }
    for (int j = Head[S]; j; j = nxt[j]) {
      f[j]++;
    }
    for (int x = n1 + 1; x <= n; x++) {
      for (int j = Head[x]; j; j = nxt[j]) {
        v = to[j];
        if (v == T) {
          f[j]++;
        }
      }
    }
  }
  for (int i = 0; i <= Mindegree; i++) {
    printf("%d", ans[i][0]);
    for (int j = 1; j <= ans[i][0]; j++) {
      printf(" %d", ans[i][j]);
    }
    puts("");
  }
  return 0;
}