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;
long long int power(long long int a, long long int b, long long int m) {
  if (b == 0) return 1;
  if (b == 1) return a % m;
  long long int t = power(a, b / 2, m);
  t = (t * t) % m;
  if (b & 1) t = (t * a) % m;
  return t;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) {
    return a;
  } else {
    return gcd(b, a % b);
  }
}
int main() {
  long long int n;
  cin >> n;
  vector<long long int> v[n + 1];
  vector<pair<long long int, long long int>> v1;
  for (long long int i = 0; i < n - 1; i++) {
    long long int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
    v1.push_back({x, y});
  }
  map<pair<long long int, long long int>, long long int> ans;
  long long int k = 0;
  for (long long int i = 1; i <= n; i++) {
    if (v[i].size() == 1 && (ans.find({v[i][0], i}) == ans.end()) &&
        (ans.find({i, v[i][0]}) == ans.end())) {
      ans.insert({{v[i][0], i}, k});
      k++;
    }
  }
  long long int p = n - 2;
  for (long long int i = 0; i < v1.size(); i++) {
    if (ans.find({v1[i].first, v1[i].second}) == ans.end() &&
        ans.find({v1[i].second, v1[i].first}) == ans.end()) {
      cout << p << endl;
      p--;
    } else {
      if (ans.find({v1[i].first, v1[i].second}) != ans.end()) {
        cout << ans[{v1[i].first, v1[i].second}] << endl;
      } else {
        cout << ans[{v1[i].second, v1[i].first}] << endl;
      }
    }
  }
}