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 C++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
#define rep(i, n) for(int i= 0; i < (n); i++)
using ll= long long int;
using namespace std;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
ll mod= 1e9 + 7;



int N, M;  //N頂点M辺
vector<int> visited;
vector<vector<int>> G;


int main() {
    cin >> N;
    M=N-1;
    pair<int,int>p[M];
    G = vector<vector<int>>(N); // サイズを確保する
    visited = vector<int>(N, 0); // 全頂点を「未訪問」に初期化
    for(int i=0; i<M; i++){
        int a, b;
        cin >> a >> b;
        // 頂点番号を 0-indexed とする
        a--;
        b--;
        G[a].push_back(b);
        G[b].push_back(a);
        p[i]=make_pair(a,b);
    }
    
    int ans[N+1][N+1]={};//2行目以降
    int qq[N];
    qq[0]=0;

    queue<int> que;
    int ft=0; //大体 0 なのでft=0としてカットしても良い

    que.push(ft); // ft を橙色頂点にする
    visited[ft]=1; // 初期条件 (頂点 ft を初期ノードとする)

    // BFS 開始 (キューが空になるまで探索を行う)
    while (!que.empty()) {
        int v = que.front(); // キューから先頭頂点を取り出す
        que.pop();
        int d=1;

        // v から辿れる頂点をすべて調べる
        for (int nv : G[v]) {
            if (visited[nv] != 0) continue; // すでに発見済みの頂点は探索しない
            if(qq[v]==d){d++;qq[nv]=d;ans[v][nv]=d;}else{qq[nv]=d;ans[v][nv]=d;}
            d++;

            // 新たな白色頂点 nv についてキューに追加する（距離情報を更新する場合はvisited[v]+1）
            visited[nv] = 1;
            que.push(nv);
        }
    }
    visited[0] = 0; //放置だとvisited[0]=0にならない為

  int q=0;
  rep(i,N)chmax(q,qq[i]);
  cout << q << endl;
  for(int i=0;i<M;i++){
    cout << ans[p[i].first][p[i].second] << endl;
  }

    
}