CF 638 dfs

http://codeforces.com/contest/638/problem/C

C. Road Improvement

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.

In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simultaneously for one day. Both brigades repair one road for the whole day and cannot take part in repairing other roads on that day. But the repair brigade can do nothing on that day.

Determine the minimum number of days needed to repair all the roads. The brigades cannot change the cities where they initially are.

Input

The first line of the input contains a positive integer n (2 ≤ n ≤ 200 000) — the number of cities in Berland.

Each of the next n - 1 lines contains two numbers ui, vi, meaning that the i-th road connects city ui and city vi (1 ≤ ui, vi ≤ n, ui ≠ vi).

Output

First print number k — the minimum number of days needed to repair all the roads in Berland.

In next k lines print the description of the roads that should be repaired on each of the k days. On the i-th line print first number di — the number of roads that should be repaired on the i-th day, and then di space-separated integers — the numbers of the roads that should be repaired on the i-th day. The roads are numbered according to the order in the input, starting from one.

If there are multiple variants, you can print any of them.

Examples

input

4
1 2
3 4
3 2

output

2
2 2 1
1 3

input

6
3 4
5 4
3 2
1 3
4 6

output

3
1 1 
2 2 3 
2 4 5 

Note

In the first sample you can repair all the roads in two days, for example, if you repair roads 1 and 2 on the first day and road 3 — on the second day.

题目大意：

n个城市，n-1条损坏道路，两个城市要一起修同一条道路才能修复，期间不能再去修改其他的道路，问，最多需要几天

先输出总天数。再先输出该天数所要修复的道路的个数，再输出道路的编号

思路：

反正是树，dfs一下就好了。只要注意下天数就行。

#include
    
    
     
     

using namespace std;

typedef pair
     
     
      
       P;
const int maxn = 200000 + 5;
int n;
vector 
     
     
    
    
 edge[maxn];
vector 
     
     
      
       a[maxn];
int res;

void dfs(int u, int past, int day){
    int now = 0;
    for (int i = 0; i < edge[u].size(); i++){
        int v = edge[u][i].first;
        if (v == past) continue;
        int road = edge[u][i].second;
        now++;
        if (now == day) now++;
        res = max(now, res);
        dfs(v, u, now);
        a[now].push_back(road);
    }
}

int main(){
    scanf("%d", &n);
    for (int i = 1; i < n; i++){
        int a, b;
        scanf("%d %d", &a, &b);
        edge[a].push_back(make_pair(b, i));
        edge[b].push_back(make_pair(a, i));
    }
    dfs(1, 0, 0);
    printf("%d\n", res);
    for (int i = 1; i <= res; i++){
        int l = a[i].size();
        printf("%d ", l);
        for (int j = 0; j < l; j++){
            printf("%d%c", a[i][j], j == l-1 ? '\n' : ' ');
        }
    }
    return 0;
}