Problem: You are given a tree consisting of n nodes. You want to write some labels on the tree's edges such that the following conditions hold:

  * Every label is an integer between 0 and n-2 inclusive. 
  * All the written labels are distinct. 
  * The largest value among MEX(u,v) over all pairs of nodes (u,v) is as small as possible. 



Here, MEX(u,v) denotes the smallest non-negative integer that isn't written on any edge on the unique simple path from node u to node v.

Input

The first line contains the integer n (2 ≤ n ≤ 10^5) — the number of nodes in the tree.

Each of the next n-1 lines contains two space-separated integers u and v (1 ≤ u,v ≤ n) that mean there's an edge between nodes u and v. It's guaranteed that the given graph is a tree.

Output

Output n-1 integers. The i^{th} of them will be the number written on the i^{th} edge (in the input order).

Examples

Input


3
1 2
1 3


Output


0
1


Input


6
1 2
1 3
2 4
2 5
5 6


Output


0
3
2
4
1

Note

The tree from the second sample:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
T maxm(T a, T b) {
  return (a > b) ? a : b;
}
template <typename T>
T minm(T a, T b) {
  return (a < b) ? a : b;
}
template <typename T>
T power(T e, T n) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = x * p;
    p = p * p;
    n >>= 1;
  }
  return x;
}
template <typename T>
T powerI(T e, int n) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = x * p;
    p = p * p;
    n >>= 1;
  }
  return x;
}
template <typename T>
T powerm(T e, T n, T m) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = (x * p) % m;
    p = (p * p) % m;
    n >>= 1;
  }
  return x;
}
template <typename T>
T InverseEuler(T a, T m) {
  return (a == 1 ? 1 : powerm(a, m - 2, m));
}
template <typename T>
T lcm(T a, T b) {
  return (a * (b / __gcd(a, b)));
}
template <typename T>
void outPrecision(T a) {
  std::cout << fixed << setprecision(8) << a;
}
string StringToUpper(string strToConvert) {
  std::transform(strToConvert.begin(), strToConvert.end(), strToConvert.begin(),
                 ::toupper);
  return strToConvert;
}
string StringToLower(string strToConvert) {
  std::transform(strToConvert.begin(), strToConvert.end(), strToConvert.begin(),
                 ::tolower);
  return strToConvert;
}
long long MAX = (long long)(3 * 1e18);
long long MOD = (long long)(1e9 + 7);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int t;
  long long n;
  cin >> n;
  vector<long long> *v = new vector<long long>[n];
  long long maxAd = -1;
  long long node = -1;
  vector<pair<long long, long long> > in;
  for (long long i = (0); i < (n - 1); ++i) {
    long long a, b;
    cin >> a >> b;
    in.push_back(make_pair(a - 1, b - 1));
    v[a - 1].push_back(b - 1);
    v[b - 1].push_back(a - 1);
    long long va_size = v[a - 1].size();
    long long vb_size = v[b - 1].size();
    if (va_size > maxAd) {
      maxAd = v[a - 1].size();
      node = a - 1;
    }
    if (vb_size > maxAd) {
      maxAd = v[b - 1].size();
      node = b - 1;
    }
  }
  if (maxAd > 2) {
    vector<long long> zo2;
    zo2.push_back(0);
    zo2.push_back(1);
    zo2.push_back(2);
    int others = 3;
    for (long long i = (0); i < (n - 1); ++i) {
      if ((in[i].first == node || in[i].second == node) && zo2.size() > 0) {
        cout << zo2.back() << endl;
        zo2.pop_back();
      } else {
        cout << others++ << endl;
      }
    }
  } else {
    for (long long i = (0); i < (n - 1); ++i) {
      cout << i << endl;
    }
  }
  return 0;
}