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 double PI = 3.14159265358979323846;
const long long MOD = 998244353, MAX = 1e5 + 1;
int n;
vector<int> adj[MAX], vis(MAX);
deque<int> nums;
map<pair<int, int>, int> mp;
void dfs(int node, int src) {
  vis[node]++;
  for (auto i : adj[node]) {
    if (!vis[i]) {
      if (node == src)
        mp[{min(i, node), max(i, node)}] = 0, dfs(i, src);
      else {
        if (adj[i].size() == 1)
          mp[{min(i, node), max(i, node)}] = nums.front(), nums.pop_front(),
                                   dfs(i, src);
        else
          mp[{min(i, node), max(i, node)}] = nums.back(), nums.pop_back(),
                                   dfs(i, src);
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  int a[n - 1], b[n - 1];
  for (int i = 0; i < n - 1; i++) {
    cin >> a[i] >> b[i];
    adj[a[i]].push_back(b[i]);
    adj[b[i]].push_back(a[i]);
  }
  for (int i = 1; i < n - 1; i++) nums.push_back(i);
  vector<int> v;
  for (int i = 1; i < n + 1; i++)
    if (adj[i].size() == 1) v.push_back(i);
  dfs(v.front(), v.front());
  for (int i = 0; i < n - 1; i++)
    cout << mp[{min(a[i], b[i]), max(a[i], b[i])}] << "\n";
}