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 inf = LLONG_MAX;
long long gcd(long long a, long long b) {
  if (!b)
    return a;
  else
    return gcd(b, a % b);
}
long long modexp(long long x, long long y, long long mod) {
  if (!y) return 1;
  long long z = modexp(x, y / 2, mod);
  z = z * z % mod;
  if (y % 2) z = z * x % mod;
  return z;
}
long long bits(long long x) {
  long long res = 0;
  while (x) {
    x = (x >> 1);
    res++;
  }
  return res;
}
void solve() {
  long long n;
  cin >> n;
  vector<pair<long long, long long> > e(n - 1);
  vector<long long> deg(n, 0);
  long long v = -1;
  for (long long i = (0); i < (n - 1); i++) {
    cin >> e[i].first >> e[i].second;
    e[i].first--;
    e[i].second--;
    deg[e[i].first]++;
    deg[e[i].second]++;
    if (deg[e[i].first] >= 3) {
      v = e[i].first;
    }
    if (deg[e[i].second] >= 3) {
      v = e[i].second;
    }
  }
  vector<long long> ans(n - 1, -1);
  long long cnt = 0;
  if (v != -1) {
    for (long long i = (0); i < (n - 1); i++) {
      if (cnt < 3 && e[i].first == v) {
        ans[i] = cnt++;
      }
      if (cnt < 3 && e[i].second == v) {
        ans[i] = cnt++;
      }
    }
  }
  for (long long i = (0); i < (n - 1); i++) {
    if (ans[i] < 0) {
      ans[i] = cnt++;
    }
  }
  for (long long i = (0); i < (n - 1); i++) cout << ans[i] << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(10);
  solve();
  return 0;
}
