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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long power(long long a, long long b, long long modi) {
  long long res = 1;
  while (b) {
    if (b % 2) {
      res = (res * a) % modi;
    }
    b /= 2;
    a = (a * a) % modi;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    vector<int> adj[n + 1];
    int u[n + 1];
    int v[n + 1];
    for (int i = 1; i <= n - 1; i++) {
      int x, y;
      cin >> x >> y;
      u[i] = x;
      v[i] = y;
      adj[x].push_back(y);
      adj[y].push_back(x);
    }
    vector<pair<int, int> > vec;
    for (int i = 1; i <= n; i++) {
      vec.push_back(make_pair(adj[i].size(), i));
    }
    sort(vec.begin(), vec.end());
    map<pair<int, int>, int> vis;
    int k = 0;
    for (int i = 0; i < vec.size(); i++) {
      for (int j = 0; j < vec[i].first; j++) {
        vis[{vec[i].second, adj[vec[i].second][j]}] = -1;
      }
    }
    for (int i = 0; i < vec.size(); i++) {
      for (int j = 0; j < vec[i].first; j++) {
        if (vis[{vec[i].second, adj[vec[i].second][j]}] == -1) {
          vis[{vec[i].second, adj[vec[i].second][j]}] = k;
          vis[{adj[vec[i].second][j], vec[i].second}] = k;
          k++;
        }
      }
    }
    for (int i = 1; i <= n - 1; i++) {
      cout << vis[{u[i], v[i]}] << '\n';
    }
  }
  return 0;
}