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 M = 1e9 + 7;
long long dx[] = {1, -1, 0, 0, 1, 1, -1, -1};
long long dy[] = {0, 0, 1, -1, 1, -1, 1, -1};
long long expo(long long x, long long y) {
  long long res = 1;
  x = x % M;
  while (y > 0) {
    if (y & 1) res = (1ll * res * x) % M;
    y = y >> 1;
    x = (1ll * x * x) % M;
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1;
  while (t--) {
    long long n, k;
    cin >> n;
    long long hash[n], cnt = 0;
    memset(hash, -1, sizeof hash);
    vector<pair<long long, long long>> v;
    map<long long, long long> mp;
    for (long long i = 0; i < n - 1; i++) {
      long long x, y;
      cin >> x >> y;
      mp[x]++;
      mp[y]++;
      v.push_back({x, y});
    }
    for (long long i = 0; i < v.size(); i++) {
      if (mp[v[i].first] == 1 || mp[v[i].second] == 1) {
        hash[i] = cnt++;
      }
    }
    for (long long i = 0; i < n - 1; i++) {
      if (hash[i] == -1) hash[i] = cnt++;
    }
    for (long long i = 0; i < n - 1; i++) {
      cout << hash[i] << "\n";
    }
  }
  return 0;
}
