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;
const long long N = 1e6 + 9, mod = 1e9 + 7, inf = 1e18;
vector<pair<long long, long long>> vc[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  bool f = true;
  for (int i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    vc[u].push_back({v, i});
    vc[v].push_back({u, i});
    if (vc[v].size() > 2 || vc[u].size() > 2) f = false;
  }
  if (f)
    for (int i = 0; i < n - 1; i++) cout << i << "\n";
  else {
    for (int i = 1; i <= n; i++) {
      if (vc[i].size() > 2) {
        int ed[n + 3], nd = 0;
        memset(ed, -1, sizeof ed);
        for (auto x : vc[i]) ed[x.second] = nd++;
        for (int i = 0; i < n - 1; i++)
          if (ed[i] == -1) ed[i] = nd++;
        for (int i = 0; i < n - 1; i++) cout << ed[i] << "\n";
        return 0;
      }
    }
  }
}