// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

#define F first

#define S second

#define pii pair<int, int>

#define eb emplace_back

#define all(v) v.begin(), v.end()

#define rep(i, n) for (int i = 0; i < (n); ++i)

#define rep3(i, l, n) for (int i = l; i < (n); ++i)

#define sz(v) (int)v.size()

const int inf = 1e9 + 7;

const ll INF = 1e18;

int mod = 1000000007;

#define abs(x) (x >= 0 ? x : -(x))

#define lb(v, x) (int)(lower_bound(all(v), x) - v.begin())

#define ub(v, x) (int)(upper_bound(all(v), x) - v.begin())

template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }

template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }

template<typename T> T gcd(T a, T b) { if (b == 0) return a; return gcd(b, a % b); }

template<typename T> T lcm(T a, T b) { return a / gcd(a, b) * b; }

template<typename T> T pow(T a, int b) { return b ? pow(a * a, b / 2) * (b % 2 ? a : 1) : 1; }

ll modpow(ll a, ll b, ll _mod) { return b ? modpow(a * a % _mod, b / 2, _mod) * (b % 2 ? a : 1) % _mod : 1; }

template<class T> ostream& operator << (ostream& os, const vector<T>& vec) { for (auto &vi: vec) os << vi << " "; return os; }

template<class T, class U> ostream& operator << (ostream& os, const pair<T, U>& p) { os << p.F << " " << p.S; return os; }

template<typename T >inline istream& operator >> (istream& is, vector<T>& v) { rep(j, sz(v)) is >> v[j]; return is; }

template<class T> inline void add(T &a, int b) { a += b; if (a >= mod) a -= mod; }





void solve();



int main() {

    ios::sync_with_stdio(false);

    cin.tie(0);

    cout.tie(0);

    cout << fixed << setprecision(10);



    int T;

    // cin >> T;

    T = 1;



    while (T--) {

        solve();

    }

}



const int N = 100005;

vector<vector<pii> > G(N);

vector<int> color(N), used(N);



void dfs(int v, int p, int c) {

  // cout << "v " << v << " " << p << " " << c << endl;

  used[v] = 1;

  int co = 1;

  for (pii P : G[v]) {

    int nv = P.F, e = P.S;

    if (nv == p) continue;



    if (co == c) co++;

    color[e] = co;

    // cout << nv << " " << v << " " << co << endl;

    dfs(nv, v, co);

    co++; // dfs のあと

  }

}

void solve() {

  int n;

  cin >> n;

  rep(i, n - 1) {

    int a, b;

    cin >> a >> b;

    a--; b--;

    G[a].eb(b, i);

    G[b].eb(a, i);

  }

  // 何色使うか

  // 字数最大の頂点

  int K = 0;

  rep(i, n) chmax(K, sz(G[i]));

  cout << K << endl;



  dfs(0, -1, -1);

  rep(i, n - 1) cout << color[i] << endl;

}


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

#define F first

#define S second

#define pii pair<int, int>

#define pli pair<ll, int>

#define pil pair<int, ll>

#define pll pair<ll, ll>

#define eb emplace_back

#define all(v) v.begin(), v.end()

#define rep(i, n) for (int i = 0; i < (n); ++i)

#define rep3(i, l, n) for (int i = l; i < (n); ++i)

#define sz(v) (int)v.size()

#define endl '\n'

const int inf = 1000000007;

const ll INF = 1e18;

// int mod = 998244353;

int mod = 1000000007;

#define abs(x) (x >= 0 ? x : -(x))

#define lb(v, x) (int)(lower_bound(all(v), x) - v.begin())

#define ub(v, x) (int)(upper_bound(all(v), x) - v.begin())

template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }

template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }

ll gcd(ll a, ll b) { if (b == 0) return a; return gcd(b, a % b); }

ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }

template<typename T, typename U> T pow_(T a, U b) { return b ? pow_(a * a, b / 2) * (b % 2 ? a : 1) : 1; }

ll modpow(ll a, ll b, ll _mod) { return b ? modpow(a * a % _mod, b / 2, _mod) * (b % 2 ? a : 1) % _mod : 1; }

template<class T, class U> ostream& operator << (ostream& os, const pair<T, U>& p) { os << p.F << " " << p.S; return os; }

template<class T> ostream& operator << (ostream& os, const vector<T>& vec) { rep(i, sz(vec)) { if (i) os << " "; os << vec[i]; } return os; }

template<typename T> inline istream& operator >> (istream& is, vector<T>& v) { rep(j, sz(v)) is >> v[j]; return is; }

template<class T, class T2> inline void add(T &a, T2 b) { a += b % mod; if (a >= mod) a -= mod; }

template<class T> void operator += (vector<T>& v, vector<T> v2) { rep(i, sz(v2)) v.eb(v2[i]); }



void solve();



int main() {

    ios::sync_with_stdio(false);

    cin.tie(0);

    cout.tie(0);

    cout << fixed << setprecision(12); // debuged

    int T;

    T = 1;

    while (T--) solve();

}



vector<vector<pii> > G(100005);

vector<int> e(100005, -1);

void dfs(int v) {

    int col = -1;

    for (pii p : G[v]) {

        if (e[p.S] != -1) {

            col = e[p.S];

            break;

        }

    }

    int c = 1;

    rep(i, sz(G[v])) {

        if (e[G[v][i].S] != -1) continue;

        if (c == col) ++c;

        e[G[v][i].S] = c++;

        dfs(G[v][i].F);

    }

}



void solve() {

    int n;

    cin >> n;

    rep(i, n - 1) {

        int a, b;

        cin >> a >> b;

        --a; --b;

        G[a].eb(b, i);

        G[b].eb(a, i);

    }

    

    dfs(0);

    cout << *max_element(all(e)) << endl;

    rep(i, n - 1) cout << e[i] << endl;

}


// Fastest known solution for this problem: