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;
mt19937 rnd(time(0));
template <class T>
long long upmax(T &a, T b) {
  return (b > a) ? a = b, 1 : 0;
}
template <class T>
long long upmin(T &a, T b) {
  return (b < a) ? a = b, 1 : 0;
}
vector<vector<pair<int, int>>> graph;
void solve() {
  int n;
  cin >> n;
  graph.resize(n);
  for (int i = 0, x, y; i < n - 1; i++) {
    cin >> x >> y;
    x--, y--, graph[x].push_back({y, i}), graph[y].push_back({x, i});
  }
  vector<int> ans(n - 1, -1);
  int what = 0;
  vector<bool> used(n, 0);
  queue<int> que;
  for (int i = 0; i < n; i++) {
    if (graph[i].size() == 1) que.push(i), used[i] = 1;
  }
  while (que.size()) {
    int v = que.front();
    que.pop();
    for (auto [nxt, id] : graph[v]) {
      if (ans[id] == -1) {
        ans[id] = what++;
        if (!used[nxt]) que.push(nxt), used[nxt] = 1;
      }
    }
  }
  for (auto i : ans) cout << i << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
