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;
void debug_out() { cerr << '\n'; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
const int maxn = 3e5 + 10;
int n, u[maxn], v[maxn], ind[maxn], ans[maxn];
vector<int> adj[maxn];
vector<int> leaf;
void dfs(int now, int prev) {
  bool chk = false;
  for (auto nxt : adj[now])
    if (nxt != prev) {
      dfs(nxt, now);
      chk = true;
    }
  if (!chk) {
    leaf.push_back(now);
  }
}
int main(void) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n - 1; ++i) {
    cin >> u[i] >> v[i];
    adj[u[i]].push_back(v[i]);
    adj[v[i]].push_back(u[i]);
    ++ind[u[i]];
    ++ind[v[i]];
  }
  int idx = -1;
  for (int i = 1; i <= n; ++i)
    if (ind[i] > 2) idx = i;
  if (idx == -1) {
    for (int i = 0; i < n - 1; ++i) cout << i << '\n';
    return 0;
  }
  vector<int> ve;
  for (int i = 0; i < 3; ++i) {
    dfs(adj[idx][i], idx);
    ve.push_back(leaf.back());
  }
  memset(ans, -1, sizeof(ans));
  for (int i = 1; i <= n - 1; ++i) {
    for (int j = 0; j < 3; ++j)
      if (ve[j] == u[i] || ve[j] == v[i]) ans[i] = j;
  }
  int cnt = 3;
  for (int i = 1; i <= n - 1; ++i) {
    if (ans[i] == -1) ans[i] = cnt++;
    cout << ans[i] << '\n';
  }
  return 0;
}