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 int QQ = 3e5 + 10;
const int maxN = (1 << 21) + 100;
const int maxM = 3e5 + 100;
const long long oo = 4e18;
const int INF = 1e9 + 100;
const int MOD = 1e9 + 7;
const int Base = 311;
const long double du = 1e-9;
const string YES = "YES";
const string NO = "NO";
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, 1, 0, -1};
void doc() {
  freopen(
      "solve"
      ".inp",
      "r", stdin);
  freopen(
      "solve"
      ".out",
      "w", stdout);
}
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int> > adj[n];
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].push_back({v, i});
    adj[v].push_back({u, i});
  }
  int x = 0;
  for (int i = 1; i < n; i++) {
    if ((int)adj[i].size() > (int)adj[x].size()) x = i;
  }
  vector<int> res(n - 1, -1);
  int L = 0;
  for (auto it : adj[x]) res[it.second] = L++;
  for (int i = 0; i < n - 1; i++) {
    if (res[i] == -1) res[i] = L++;
  }
  for (int i = 0; i < n - 1; i++) cout << res[i] << endl;
}
int main(int argc, char *argv[]) {
  srand(time(nullptr));
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  solve();
}