Problem:
Score : 400 points

Problem Statement

Given is a tree G with N vertices.
The vertices are numbered 1 through N, and the i-th edge connects Vertex a_i and Vertex b_i.

Consider painting the edges in G with some number of colors.
We want to paint them so that, for each vertex, the colors of the edges incident to that vertex are all different.

Among the colorings satisfying the condition above, construct one that uses the minimum number of colors.

Constraints

2 \le N \le 10^5

1 \le a_i \lt b_i \le N

All values in input are integers.

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
a_2 b_2
\vdots
a_{N-1} b_{N-1}

Output

Print N lines.

The first line should contain K, the number of colors used.

The (i+1)-th line (1 \le i \le N-1) should contain c_i, the integer representing the color of the i-th edge, where 1 \le c_i \le K must hold.

If there are multiple colorings with the minimum number of colors that satisfy the condition, printing any of them will be accepted.

Sample Input 1

3
1 2
2 3

Sample Output 1

2
1
2

Sample Input 2

8
1 2
2 3
2 4
2 5
4 7
5 6
6 8

Sample Output 2

4
1
2
3
4
1
1
2

Sample Input 3

6
1 2
1 3
1 4
1 5
1 6

Sample Output 3

5
1
2
3
4
5

Solution in Python (3.4.3) (judge verdict: Wrong Answer)

import sys
sys.setrecursionlimit(500000)
N = int(input())
E = [[] for _ in range(N+1)]
for i in range(N-1):
    a, b = map(int, input().split())
    E[a].append((b, i))
    E[b].append((a, i))
K = max(len(e) for e in E)
print(K)
Ans = [-1] * (N-1)
def dfs(v=1, p=0, p_col=-1):
    col = 1
    for u, idx in E[v]:
        if u!=p: #木構造の逆向きの矢印を除くための条件
            if col == p_col:
                col += 1
            Ans[idx] = col
            print(p_col)
            dfs(u, v, col) #u;調べる頂点 v;直前に調べた辺で、調べる頂点と結ばれている頂点　col;直前に調べた辺の色番号
            col += 1 #最深部まで調べたら デフォルトの色=最後に調べた辺の色+1とする
            #最深部に繋がった辺の色+1してやれば、次の辺の色は必ず条件を満たしてくれる

dfs()
print("\n".join(map(str, Ans)))