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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
long long exp(long long n, long long m, long long md) {
  long long a = 1;
  while (m > 0) {
    if (m & 1) a = (a * n % md) % md;
    m /= 2;
    n = (n * n) % md;
  }
  return a % md;
}
void printarray(int a[], int n) {
  for (int i = 0; i < n; i++) cout << a[i] << " ";
  cout << "\n";
}
void printvector(vector<int> v) {
  for (auto x : v) cout << x << " ";
  cout << "\n";
}
int main() {
  int n;
  cin >> n;
  vector<int> adj[n + 1];
  vector<pair<int, int>> v;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
    v.push_back({a, b});
  }
  int x = 0;
  int index = 0;
  for (int i = 1; i <= n; i++) {
    int sz = adj[i].size();
    x = max(x, sz);
    if (sz >= 3) index = i;
  }
  if (x >= 3) {
    int t = 3;
    int count1 = 3;
    int count2 = 0;
    for (int i = 0; i < n - 1; i++) {
      if (t > 0 && (v[i].first == index || v[i].second == index)) {
        cout << count2++ << "\n";
        t--;
      } else
        cout << count1++ << "\n";
    }
  } else {
    for (int i = 0; i < n - 1; i++) {
      cout << i << "\n";
    }
  }
}
