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 int MAXN = 1000005;
const int SQRTN = 1003;
const int LOGN = 22;
const double PI = acos(-1);
const int INF = 1e9;
int MOD = 1000000007;
const int FMOD = 998244353;
const double eps = 1e-9;
long long gcd(long long a, long long b) {
  if (a == 0) return (b);
  return (gcd(b % a, a));
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
void solve() {
  long long n;
  cin >> n;
  vector<long long> mp[n + 5];
  long long c1, c2;
  for (int i = 1; i < n; i++) {
    cin >> c1 >> c2;
    mp[c1].push_back(i);
    mp[c2].push_back(i);
  }
  pair<long long, long long> mx(0, 0);
  for (int i = 1; i <= n; i++) {
    mx = max(mx, {mp[i].size(), i});
  }
  int cur = 0;
  long long ans[n + 5];
  memset(ans, -1, sizeof(ans));
  for (int i = 0; i < mx.first; i++) {
    ans[mp[mx.second][i]] = cur++;
  }
  for (int i = 1; i < n; i++) {
    if (ans[i] == -1) ans[i] = cur++;
    cout << ans[i] << endl;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return (0);
}