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 N = 2e5 + 5;
const long long N1 = 100005;
const long long mod = 1000000007;
long long pow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    b /= 2;
    a = (a * a) % mod;
  }
  return ans;
}
long long n, t, k, m;
string s;
vector<long long> adj[N1];
long long ans[N1];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    adj[a].push_back(i);
    adj[b].push_back(i);
    ans[i] = -1;
  }
  pair<long long, long long> mx(0, 0);
  for (long long i = 1; i <= n; i++) {
    mx = max(mx, make_pair((long long)adj[i].size(), i));
  }
  long long curr = 0;
  for (long long i : adj[mx.second]) {
    ans[i] = curr++;
  }
  for (long long i = 1; i <= n - 1; i++) {
    if (ans[i] == -1) ans[i] = curr++;
    cout << ans[i] << endl;
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
  return 0;
}