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 mod = 1000000007;
const int N = 1e5 + 9;
vector<int> v[N];
bool visited[N];
int c = 0;
map<pair<int, int>, int> m;
void dfs(int x, int parent) {
  visited[x] = true;
  bool b = false;
  for (int i = 0; i < v[x].size(); i++) {
    if (!visited[v[x][i]]) {
      b = true;
      dfs(v[x][i], x);
    }
  }
  if (!b) m[{parent, x}] = c++;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int x;
  cin >> x;
  int l[x - 1], r[x - 1];
  for (int i = 0; i < x - 1; i++) {
    cin >> l[i] >> r[i];
    v[l[i]].push_back(r[i]);
    v[r[i]].push_back(l[i]);
  }
  dfs(l[0], l[0]);
  for (int i = 0; i < x - 1; i++) {
    if (m.find({l[i], r[i]}) != m.end())
      cout << m[{l[i], r[i]}] << '\n';
    else if (m.find({r[i], l[i]}) != m.end())
      cout << m[{r[i], l[i]}] << '\n';
    else
      cout << c++ << '\n';
  }
  return 0;
}