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 int mod = 1e9 + 7;
const int inf = 1e9;
const int N = 1e6;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  int a[n - 1][2];
  vector<int> adj[n + 1];
  for (int i = 0; i < n - 1; i++) {
    cin >> a[i][0] >> a[i][1];
    adj[a[i][0]].push_back(a[i][1]);
    adj[a[i][1]].push_back(a[i][0]);
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if ((int)adj[i].size() == 1) cnt++;
  }
  if (cnt == 2) {
    for (int i = 0; i < n - 1; i++) cout << i << "\n";
    return 0;
  }
  cnt = 0;
  vector<int> x;
  for (int i = 3; i < n - 1; i++) x.push_back(i);
  for (int i = 0; i < n - 1; i++) {
    int u = a[i][0];
    int v = a[i][1];
    if ((int)adj[u].size() == 1 || (int)adj[v].size() == 1) {
      if (cnt == 0 || cnt == 1 || cnt == 2) {
        cout << cnt << "\n";
        cnt++;
      } else {
        cout << x.back() << "\n";
        x.pop_back();
      }
    } else {
      cout << x.back() << "\n";
      x.pop_back();
    }
  }
  return 0;
}