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;
const long long int MAX_N = 1e5 + 7, delta = 10343;
long long int cnt[MAX_N];
map<pair<long long int, long long int>, long long int> mp;
vector<long long int> adj[MAX_N];
int32_t main() {
  long long int n;
  cin >> n;
  for (long long int i = 1; i < n; i++) {
    long long int a, b;
    cin >> a >> b;
    cnt[a]++, cnt[b]++;
    adj[a].push_back(b), adj[b].push_back(a);
    mp[{a, b}] = mp[{b, a}] = i;
  }
  long long int ans[3] = {}, idx = 0;
  for (long long int i = 1; i <= n; i++) {
    if (cnt[i] == 1) {
      ans[idx++] = i;
      if (idx > 2) break;
    }
  }
  if (idx <= 2) {
    for (long long int i = 0; i < n - 1; i++) cout << i << endl;
    return 0;
  }
  long long int sz = 3;
  long long int j[MAX_N] = {};
  j[mp[{ans[0], adj[ans[0]][0]}]] = 1;
  j[mp[{ans[1], adj[ans[1]][0]}]] = 2;
  j[mp[{ans[2], adj[ans[2]][0]}]] = 3;
  for (long long int i = 1; i < n; i++) {
    if (j[i])
      cout << j[i] - 1 << endl;
    else
      cout << sz++ << endl;
  }
  return 0;
}
