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 power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
bool dfs(long long v, long long vis[], vector<long long> graph[]) {
  vis[v] = 1;
  cout << v << " ";
  for (auto i : graph[v]) {
    if (!vis[i]) {
      vis[i] = 1;
      dfs(i, vis, graph);
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> graph[n + 1];
    memset(graph, 0, sizeof(graph));
    long long vis[n + 1];
    memset(vis, 0, sizeof(vis));
    pair<long long, long long> edge[n - 1];
    for (int i = 0; i < n - 1; i++) {
      long long a, b;
      cin >> a >> b;
      edge[i] = {a, b};
      graph[a].push_back(b);
      graph[b].push_back(a);
    }
    long long size = 0;
    long long node = 0;
    for (int i = 1; i <= n; i++) {
      if (graph[i].size() > size) {
        size = graph[i].size();
        node = i;
      }
    }
    if (size >= 3) {
      int k = 0, j = 3;
      for (int i = 0; i < n - 1; i++) {
        if ((edge[i].first == node || edge[i].second == node) && k < 3) {
          cout << k << endl;
          k++;
        } else {
          cout << j << endl;
          j++;
        }
      }
    } else {
      for (int i = 0; i < n - 1; i++) {
        cout << i << endl;
      }
    }
  }
}