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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.IOException;
import java.io.InputStream;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.NoSuchElementException;
import java.util.Queue;

class Scanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        } else {
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }

    private int readByte() {
        return hasNextByte() ? buffer[ptr++] : -1;
    }

    private boolean isPrintableChar(int c) {
        return 33 <= c && c <= 126;
    }

    private void skipUnprintable() {
        while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
    }

    public boolean hasNext() {
        skipUnprintable();
        return hasNextByte();
    }

    public String next() {
        if (!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while (isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    public int nextInt() {
        return (int) nextLong();
    }

    public long nextLong() {
        if (!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while (true) {
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            } else if (b == -1 || !isPrintableChar(b)) {
                return minus ? -n : n;
            } else {
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
}


public class Main {
	private static long toIndex(int n, int a, int b) {
		return n * a + b;
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner();
	    int n = sc.nextInt();
	    Digraph g = new Digraph(n);
	    Map<Long, Long> edgeMap = new HashMap<>();
	    List<Integer[]> edges = new LinkedList<>();

	    for (int i = 0; i < n - 1; ++i) {
	    	int a = sc.nextInt() - 1;
	    	int b = sc.nextInt() - 1;
	    	g.addEdge(a, b);
	    	g.addEdge(b, a);
	    	edges.add(new Integer[] {a, b});
	    }

		Queue<Integer> q = new LinkedList<Integer>();
		int[] parentColor = new int[n];
		Arrays.fill(parentColor, 0);
//		int[] colors = new int[n];
//    	Arrays.fill(colors, 0);
//		colors[0] = 1;
		q.add(0);
		parentColor[0] = -1;
		int mx = 0;
		while (!q.isEmpty())
		{
			int v = q.poll();
			int color = 1;
			for (int node : g.adj(v)) {
				if (parentColor[node] == 0) {
					if (color == parentColor[v]) color += 1;
					g.setColor(v, node, color);
					parentColor[node] = color;
					mx = Math.max(mx, color);
					color += 1;
					q.add(node);
				}
			}
		}

//		for (int c : colors) {
//			System.out.println(c);
//		}
		System.out.println(mx);
		for (Integer[] edge : edges) {
			System.out.println(g.color(edge[0], edge[1]));
		}

	}


	public static class Digraph {
		private int V;
		private HashSet<Integer>[] adj;
	    public Map<Integer, Integer> edgeColor;

		public Digraph(int V)
		{
			this.V = V;
			initializeAdj(V);
			edgeColor = new HashMap<>();
		}

		public void setColor(int a, int b, int color) {
			edgeColor.put(V * Math.max(a, b) + Math.min(a, b), color);
		}

		public int color(int a, int b) {
			int key = V * Math.max(a, b) + Math.min(a, b);
			if (edgeColor.containsKey(key)) {
				return edgeColor.get(key);
			} else {
				return 0;
			}
		}

		void initializeAdj(int V) {
			this.adj = new HashSet[V];
			for (int v = 0; v < V; ++v)
				adj[v] = new HashSet<Integer>();
		}

		void addEdge(int v, int w)
		{
			adj[v].add(w);
		}

		public HashSet<Integer> adj(int v)
		{
			return this.adj[v];
		}

		public int V()
		{
			return this.V;
		}

		public int E()
		{
			return Arrays.stream(adj)
				.mapToInt(e -> e.size())
				.sum();
		}

		public Digraph reverse()
		{
			Digraph g = new Digraph(this.V());
			for (int v = 0; v < this.V(); ++v)
			{
				for (int w : this.adj(v))
				{
					g.addEdge(w, v);
				}
			}
			return g;
		}
		void show()
		{
			for (int v = 0; v < this.V; ++v)
				for (int w : this.adj(v))
					System.out.println(v + "-" + w);
		}
	}
}