# Original solution
#!/usr/bin/env pypy3

import sys
# import math
# import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
from collections import deque                # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
# from collections import defaultdict          # subclass of dict. defaultdict(facroty)
# from collections import Counter              # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
# from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
# from heapq import nlargest, nsmallest        # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
# from itertools import product                # product(iter, repeat=n)
# from itertools import accumulate             # accumulate(iter[, f])
# from functools import reduce                 # reduce(f, iter[, init])
# from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
# from fractions import gcd                    # for Python3.4.3


def main():
    mod = 10000007                  # 10^9+7
    inf = float('inf')
    input = sys.stdin.readline      # 改行文字が残ることに注意
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def ii():  return int(eval(input()))
    def mi():  return list(map(int, input().rstrip().split()))
    def lmi(): return list(map(int, input().rstrip().split()))
    def li():  return list(input().rstrip())
    
    
    def bfs_tree_traverse(start):
        paths = deque([[start]])
        max_branch = 0
        while paths:
            # print(paths)
            current_path = paths.popleft()
            u = current_path[-1]
            previous = current_path[-2] if len(current_path)>=2 else -1
            previous_color = color[current_path[-2]][u] if len(current_path)>=2 else 0
            count = 0    # 現在の頂点 u から伸びるリンクの数を数える
            current_color = 0
            for v in adj[u]:
                count += 1    # 親へのリンクも含めて数えている
                if v != previous:
                    tmp = current_path[:]
                    tmp.append(v)
                    paths.append(tmp)
                    current_color += 1
                    if current_color == previous_color:
                        current_color += 1
                    color[u][v] = current_color
                    color[v][u] = current_color
            max_branch = max(max_branch, count)
        return max_branch

    n = ii()
    adj = [[] for _ in range(n)]
    L = [list([int(x)-1 for x in input().rstrip().split()]) for _ in range(n-1)]
    for pair in L:
        a, b = pair
        adj[a].append(b)
        adj[b].append(a)
    color = [[None] * n for _ in range(n)]
    
    print((bfs_tree_traverse(0)))
    for pair in L:
        a, b = pair
        print((color[a][b]))

if __name__ == "__main__":
    main()

# Optimized solution
#!/usr/bin/env pypy3

import sys
# import math
# import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
from collections import deque                # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
# from collections import defaultdict          # subclass of dict. defaultdict(facroty)
# from collections import Counter              # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
# from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
# from heapq import nlargest, nsmallest        # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
# from itertools import product                # product(iter, repeat=n)
# from itertools import accumulate             # accumulate(iter[, f])
# from functools import reduce                 # reduce(f, iter[, init])
# from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
# from fractions import gcd                    # for Python3.4.3


def main():
    mod = 10000007                  # 10^9+7
    inf = float('inf')
    input = sys.stdin.readline      # 改行文字が残ることに注意
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def ii():  return int(eval(input()))
    def mi():  return list(map(int, input().rstrip().split()))
    def lmi(): return list(map(int, input().rstrip().split()))
    def li():  return list(input().rstrip())
    
    
    def bfs_tree_traverse(start):
        paths = deque([[start]])
        max_branch = 0
        while paths:
            # print(paths)
            current_path = paths.popleft()
            u = current_path[-1]
            previous = current_path[-2] if len(current_path)>=2 else -1
            previous_color = color[current_path[-2]][u] if len(current_path)>=2 else 0
            count = 0    # 現在の頂点 u から伸びるリンクの数を数える
            current_color = 0
            for v in adj[u]:
                count += 1    # 親へのリンクも含めて数えている
                if v != previous:
                    # tmp = current_path[:]
                    tmp = current_path[-1:]    # 定数倍高速化への道のり
                    tmp.append(v)
                    paths.append(tmp)
                    current_color += 1
                    if current_color == previous_color:
                        current_color += 1
                    color[u][v] = current_color
                    # color[v][u] = current_color    # 定数倍高速化への道のり
            max_branch = max(max_branch, count)
        return max_branch

    n = ii()
    adj = [[] for _ in range(n)]
    L = [list([int(x)-1 for x in input().rstrip().split()]) for _ in range(n-1)]
    for pair in L:
        a, b = pair
        adj[a].append(b)
        adj[b].append(a)
    color = [[None] * n for _ in range(n)]
    
    print((bfs_tree_traverse(0)))
    for pair in L:
        a, b = pair
        print((color[a][b]))

if __name__ == "__main__":
    main()