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;
void solve() {
  int n;
  cin >> n;
  unordered_map<int, int> dcount;
  vector<pair<int, int>> edge_list;
  for (int i = 0; i < n - 1; ++i) {
    int x, y;
    cin >> x >> y;
    edge_list.push_back({x, y});
    ++dcount[x];
    ++dcount[y];
  }
  int mvalue = -1;
  int mkey = -1;
  for (auto d : dcount) {
    if (d.second > mvalue) {
      mkey = d.first;
      mvalue = d.second;
    }
  }
  if (mvalue >= 3) {
    int md_count = 0;
    int o_count = 3;
    for (auto vi : edge_list) {
      if (vi.first == mkey || vi.second == mkey) {
        if (md_count <= 2) {
          cout << md_count << "\n";
          ++md_count;
        } else {
          cout << o_count << "\n";
          ++o_count;
        }
      } else {
        cout << o_count << "\n";
        ++o_count;
      }
    }
  } else {
    int count = 0;
    for (auto vi : edge_list) {
      cout << count << "\n";
      ++count;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
