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>
#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream& operator<<(ostream& cout, vector<T> V) {
  cout << "[ ";
  for (auto v : V) cout << v << ' ';
  return cout << ']';
}
template <class L, class R>
ostream& operator<<(ostream& cout, pair<L, R> P) {
  return cout << '(' << P.first << ',' << P.second << ')';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  vector<vector<pair<long long int, long long int>>> conn(n);
  for (long long int i = 0; i < (n - 1); i++) {
    long long int x;
    cin >> x;
    long long int y;
    cin >> y;
    x--, y--;
    conn[x].push_back({y, i});
    conn[y].push_back({x, i});
  }
  vector<long long int> l;
  for (long long int i = 0; i < (n); i++)
    if (conn[i].size() == 1) l.push_back(i);
  if (l.size() == 2) {
    for (long long int i = 0; i < (n - 1); i++) cout << i << '\n';
    return 0;
  }
  vector<long long int> ans(n - 1, -1);
  ans[conn[l[0]][0].second] = 0;
  ans[conn[l[1]][0].second] = 1;
  ans[conn[l[2]][0].second] = 2;
  long long int curr = 3;
  for (long long int i = 0; i < (n - 1); i++) {
    if (ans[i] != -1) continue;
    ans[i] = curr;
    curr++;
  }
  for (long long int i = 0; i < (n - 1); i++) cout << ans[i] << '\n';
  cerr << "Time : " << 1000 * (long double)clock() / (long double)CLOCKS_PER_SEC
       << "ms\n";
  ;
}
