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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  map<pair<long long, long long>, long long> ans;
  map<long long, long long> ff;
  map<long long, vector<long long> > edges;
  vector<pair<long long, long long> > ve;
  long long u, v;
  for (long long i = 0; i < n - 1; i++) {
    cin >> u >> v;
    ve.push_back(make_pair(u, v));
    ff[u]++;
    ff[v]++;
    edges[u].push_back(v);
    edges[v].push_back(u);
    pair<long long, long long> temp = make_pair(u, v);
    ans[temp] = -1;
  }
  priority_queue<long long, vector<long long>, greater<long long> > q;
  for (long long i = 0; i < n - 1; i++) q.push(i);
  multimap<long long, long long, greater<long long> > f;
  for (auto x : ff) f.insert(make_pair(x.second, x.first));
  for (auto x : f) {
    long long node = x.second;
    for (auto y : edges[node]) {
      if (!q.empty()) {
        pair<long long, long long> t1, t2;
        t1 = make_pair(node, y);
        t2 = make_pair(y, node);
        if (ans[t1] == -1) {
          ans[t1] = q.top();
          q.pop();
        } else if (ans[t2] == -1) {
          ans[t2] = q.top();
          q.pop();
        }
      }
    }
  }
  for (auto x : ve) cout << ans[x] << "\n";
}
