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;
long long int add(long long int x, long long int y) {
  return (x % 1000000007 + y % 1000000007) % 1000000007;
}
long long int mul(long long int x, long long int y) {
  return ((x % 1000000007) * (y % 1000000007)) % 1000000007;
}
long long int sub(long long int x, long long int y) {
  return (x - y + 1000000007) % 1000000007;
}
long long int ans[100005];
std::vector<long long int> adj[100005];
long long int vis[100005];
long long int dis[100005];
void dfs(long long int node, long long int x) {
  vis[node] = 1;
  ans[node] = 1;
  dis[node] = x;
  for (auto i : adj[node])
    if (!vis[i]) {
      dfs(i, x + 1);
      ans[node] += ans[i];
    }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    long long int n;
    cin >> n;
    std::vector<pair<long long int, long long int>> v;
    for (int i = 0; i < n - 1; i++) {
      long long int x, y;
      cin >> x >> y;
      v.push_back({x, y});
      adj[x].push_back(y);
      adj[y].push_back(x);
    }
    dfs(1, 0);
    std::vector<pair<long long int, long long int>> iu;
    for (int i = 0; i < n - 1; i++) {
      long long int p = v[i].first;
      long long int q = v[i].second;
      long long int op;
      if (dis[p] > dis[q]) {
        op = ans[p] * (n - ans[p]);
      } else
        op = ans[q] * (n - ans[q]);
      iu.push_back({op, i});
    }
    sort((iu).begin(), (iu).end());
    long long int final[n + 3];
    long long int cnt = 0;
    for (int i = 0; i < n - 1; i++) {
      final[iu[i].second] = cnt;
      cnt++;
    }
    for (int i = 0; i < n - 1; i++) cout << final[i] << '\n';
  }
  return 0;
}
