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 big = 1000000007;
unsigned long long gcd(unsigned long long a, unsigned long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<vector<int>> adj(n + 1);
  vector<pair<int, int>> edge(n - 1, make_pair(0, 0));
  for (int i = 0; i < n - 1; i += 1) {
    cin >> edge[i].first >> edge[i].second;
    adj[edge[i].first].push_back(edge[i].second);
    adj[edge[i].second].push_back(edge[i].first);
  }
  vector<bool> mark(n + 1, false);
  vector<pair<int, int>> leaf(0);
  for (int i = 1; i <= n; i += 1) {
    if (adj[i].size() == 1 && !mark[i]) {
      mark[i] = true;
      mark[adj[i][0]] = true;
      leaf.push_back(make_pair(i, adj[i][0]));
    }
  }
  pair<int, int> temp;
  vector<int> ans(n - 1, -1);
  int i = 0;
  for (i = 0; i < leaf.size() && i < 3; i += 1) {
    temp = make_pair(leaf[i].second, leaf[i].first);
    for (int j = 0; j < n - 1; j += 1) {
      if (edge[j] == leaf[i] || edge[j] == temp) {
        ans[j] = i;
        break;
      }
    }
  }
  for (int x = 0; x < n - 1; x += 1) {
    if (ans[x] != -1)
      cout << ans[x];
    else
      cout << i, i += 1;
    cout << '\n';
  }
  return 0;
}