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 long long mod = 1e9 + 7;
const long long mx = 2e5 + 10;
int posx[] = {1, -1, 0, 0};
int posy[] = {0, 0, 1, -1};
map<pair<int, int>, int> ans, vis;
bool visited[mx];
vector<int> tree[mx], adj(mx);
int curr;
void Dfs(int src) {
  visited[src] = true;
  for (auto x : tree[src]) {
    if (!visited[x]) {
      Dfs(x);
    }
    if (adj[x] == 1 && !vis[{src, x}]) {
      ans[{x, src}] = curr;
      ans[{src, x}] = curr++;
      vis[{src, x}] = true;
      vis[{x, src}] = true;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1, n, k, m, a, b, c, d;
  while (t--) {
    cin >> n;
    vector<pair<int, int> > ve;
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      ve.push_back({u, v});
      tree[u].push_back(v);
      tree[v].push_back(u);
      adj[u]++, adj[v]++;
    }
    Dfs(1);
    for (auto x : ve) {
      int u = x.first, v = x.second;
      if (vis[{u, v}])
        cout << ans[{u, v}] << endl;
      else
        cout << curr++ << endl;
    }
  }
  return 0;
}