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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma GCC optimize("O3")
#define _CRT_SECURE_NO_WARNINGS
#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define M 1000000007
#define M2 998244353
#define ll long long
#define pll pair<long, long>
#define REP(i, a, b) for (ll i = a; i < b; i++)
#define REPI(i, a, b) for (ll i = b - 1; i >= a; i--)
#define ff first
#define ss second
#define pb push_back
#define db pop_back
#define mp make_pair
#define mt make_tuple
#define g(a, b) get<a>(b)
#define INF (ll)1e18 + 100
#define vl vector<ll>
#define vi vector<int>
#define vll vector<pair<ll, ll>>
#define vii vector<pair<int, int>>
#define all(v) v.begin(),v.end()
#define bset(a, p) ((a) | (1ll << (p)))
#define bchk(a, p) ((a) & (1ll << (p)))
#define bxor(a, p) ((a) ^ (1ll << (p)));
#define brem(a, p) (bchk(a, p) ? (bxor(a, p)) : (a))
/*SOME BITMASK KNOWLEDGE
1)x & (x - 1):sets the last one bit of x to zero
power of two exactly when x & (x − 1) = 0.
2)x & -x:sets all the one bits to zero, except last one bit
3)x | (x - 1):inverts all the bits after the last one bit*/
#define endl "\n"
#define p0(a) cout << a << " "
#define p1(a) cout << a << endl
#define p2(a, b) cout << a << " " << b << endl
#define p3(a, b, c) cout << a << " " << b << " " << c << endl
#define p4(a, b, c, d) cout << a << " " << b << " " << c << " " << d << endl

#define o_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
#define o_setll tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update>
//member functions :
//1. order_of_key(k) : number of elements strictly lesser than k
//2. find_by_order(k) : k-th element in the set

ll modI(ll a, ll m);
ll gcd(ll a, ll b);
ll powM(ll x, unsigned ll y, unsigned ll m);
void pairsort(int a[], int b[], int n);
void pairsortll(ll a[],ll b[],ll n);
ll logint(ll x,ll y);
 
 
ll gcd(ll x,ll y)
{
    if(x==0) return y;
    return gcd(y%x,x);
}
 
ll powM(ll x,ll y,ll m)
{
    if(y==0) return 1;
    ll p=powM(x,y/2,m)%m;
    p=(p*p)%m;
    return (y%2==0)?p:(x*p)%m;
}
 
ll modI(ll a, ll m)
{
   return powM(a, m-2, m);
}
 
void pairsort(int a[],int b[],int n)
{
    pair<int,int> v[n];
    REP(i,0,n)
    {
        v[i].ff=a[i];
        v[i].ss=b[i];
    }
    sort(v,v+n);
    REP(i,0,n)
    {
        a[i]=v[i].ff;
        b[i]=v[i].ss;
    }
}
 
void pairsortll(ll a[],ll b[],ll n)
{
    pair<ll,ll> v[n];
    REP(i,0,n)
    {
        v[i].ff=a[i];
        v[i].ss=b[i];
    }
    sort(v,v+n);
    REP(i,0,n)
    {
        a[i]=v[i].ff;
        b[i]=v[i].ss;
    }
}
 
ll logint(ll x,ll y)
{
    ll ans=0;
    ll a=1;
    for(ll i=0;i<=x;i++)
    {
        if(x<a)
        {
            return ans;
        }
        ans++;
        a*=y;
    }
    return -1;
}
int col=0; int n;
const int N = 1e5+5;
vector<int>v[N];
vector<int>deg(N);
vector<pair<int,int> >g;
vector<pair<int,int>>edge;
set<int>occ[N];
map<pair<int,int>,int>m;
void dfs(int ve, int p){
    for(auto x:v[ve]){
        int z = x, y = ve;
        if(z < y){swap(y,z);}
        if(m[{y,z}]){continue;}
        for(ll i = (p+1)%(col+1); i != p; i++){
            if(i > col){i %= (col+1);}
            if(!i){i++;}
            auto it = occ[x].find(i);
            //auto itr = occ[ve].find(i);
            if(it == occ[x].end()){occ[ve].insert(i); occ[x].insert(i); 
            m[{y,z}] = i;
            dfs(x, i); break;}
        }
    }
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    //Hey, You
    // Don't practise these
    // easy questions.
     cin>>n;
     REP(i,0,n-1)
    {
        int x,y;
        cin>>x>>y;
        if(x > y){swap(x,y);}
        edge.pb({x,y});
        v[x].pb(y);
        v[y].pb(x);
        deg[x]++; deg[y]++;
    }  
    REP(i,1,n+1){
        g.pb({deg[i],i});
    }
    sort(all(g));
    col = g[n-1].ff;
    dfs(g[n-1].ss,0);
    cout<<col<<"\n";
    REP(i,0,n-1){
        cout<<m[{edge[i].ff,edge[i].ss}]<<"\n";
    }
}