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: Accepted)

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using ld = long double;

using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vs = vector<string>;

typedef pair<ll, ll> P;

#define bit(n) (1LL << (n))

//#define int long long

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

#define rep(i, n) for (ll i = 0; i < n; i++)
#define REP(i, n) for (ll i = 1; i < n; i++)

#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define FORm(i, m) for (auto i = m.begin(); i != m.end(); i++)

template <class T>
inline void chmax(T& a, T b) {
  a = std::max(a, b);
}
template <class T>
inline void chmin(T& a, T b) {
  a = std::min(a, b);
}

#define mod (ll)(1e9 + 7)
#define INF LLONG_MAX

template <ll ModVal>
struct ModInt {
  ll x;

  ModInt(ll _x = 0) : x((_x % ModVal + ModVal) % ModVal) {
  }

  ModInt operator-() const {
    return ModInt(-x);
  }
  ModInt& operator+=(const ModInt a) {
    x += a.x;
    if (x >= ModVal)
      x -= ModVal;
    return *this;
  }
  ModInt& operator-=(const ModInt a) {
    x = x + ModVal - a.x;
    if (x >= ModVal)
      x -= ModVal;
    return *this;
  }
  ModInt& operator*=(const ModInt a) {
    x *= a.x;
    x %= ModVal;
    return *this;
  }

  ll ext_gcd(ll a, ll b, ll& x, ll& y) {
    if (b == 0) {
      x = 1;
      y = 0;
      return a;
    }
    ll tmp = a / b;
    ll d = ext_gcd(b, a - b * tmp, y, x);
    y -= tmp * x;
    return d;
  }

  // 逆元
  ModInt inv(const ModInt a) {
    ll u, v;
    ext_gcd(a.x, ModVal, u, v);
    return ModInt(u);
  }

  ModInt& operator/=(const ModInt a) {
    return (*this) *= inv(a);
  }

  ModInt operator+(const ModInt a) const {
    ModInt retval(*this);
    return retval += a;
  }
  ModInt operator-(const ModInt a) const {
    ModInt retval(*this);
    return retval -= a;
  }
  ModInt operator*(const ModInt a) const {
    ModInt retval(*this);
    return retval *= a;
  }
  ModInt operator/(const ModInt a) const {
    ModInt retval(*this);
    return retval /= a;
  }

  ModInt pow(ll n) {
    ModInt ans(1);
    while (n) {
      if (n & 1)
        ans = ans * x;
      *this = (*this) * (*this);
      n = n >> 1;
    }
    return ans;
  }

  constexpr const ll& value() {
    return this->x;
  }
};

template <ll ModVal>
ostream& operator<<(ostream& os, const ModInt<ModVal>& a) {
  os << a.x;
  return os;
}

using mint = ModInt<mod>;

template <typename T>
class Combination {
 public:
  Combination(ll _max_n) : max_n(_max_n), factional(max_n + 1), inv(max_n + 1) {
    factional[0] = 1;
    inv[0] = 1;
    for (ll i = 0; i < max_n; i++) {
      factional[i + 1] = factional[i] * (i + 1); // n!(mod M)
      inv[i + 1] = inv[i] / (i + 1);             // k!^(M-2) (mod M)
    }
  }

  // nCk
  T choose(ll n, ll k) {
    if (n == 0 && k == 0)
      return 1;
    if (n < k || n < 0)
      return 0;
    T tmp = inv[n - k] * inv[k];
    return tmp * factional[n];
  }

 private:
  const ll max_n;
  std::vector<T> factional;
  std::vector<T> inv;
};

using Comb = Combination<mint>;

ll n;
map<ll, vector<P>> m;
vll ans;

void dfs(ll index, ll parent_color) {
  ll c = 1;
  auto l = m[index];

  FORm(it, l) {
    if (ans[it->first]) {
      continue;
    }
    if (c == parent_color) {
      c++;
    }
    ans[it->first] = c++;
    dfs(it->second, ans[it->first]);
  }
}

signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);

  cin >> n;
  ans = vll(n - 1);

  rep(i, n - 1) {
    ll a, b;
    cin >> a >> b;
    m[a].push_back(make_pair(i, b));
    m[b].push_back(make_pair(i, a));
  }

  dfs(1, 0);

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

  rep(i, n - 1) {
    cout << ans[i] << endl;
  }

  return 0;
}