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 int MOD = 1e9 + 7;
long long power(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans *= a;
    a = a * a;
    b >>= 1;
  }
  return ans;
}
long long powm(long long a, long long b) {
  a %= MOD;
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % MOD;
    a = (a * a) % MOD;
    b >>= 1;
  }
  return ans;
}
void IO() {}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  IO();
  ;
  int n;
  cin >> n;
  vector<pair<int, int> > graph[n + 1];
  vector<int> degree(n + 1), ans(n + 1, -1);
  int res = 0;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    graph[u].push_back({v, i});
    graph[v].push_back({u, i});
    degree[u]++;
    degree[v]++;
  }
  for (int u = 1; u <= n; u++) {
    if (degree[u] != 1) continue;
    for (auto v : graph[u]) {
      if (ans[v.second] == -1) {
        ans[v.second] = res;
        res++;
      }
    }
  }
  for (int u = 1; u <= n; u++) {
    for (auto v : graph[u]) {
      if (ans[v.second] == -1) {
        ans[v.second] = res;
        res++;
      }
    }
  }
  for (int i = 1; i < n; i++) cout << ans[i] << "\n";
  return 0;
}
