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;
const int N = 2000 * 100 + 5;
bool isPrime(int a) {
  for (int i = 2; i * i <= a; ++i)
    if (a % i == 0) return false;
  return true;
}
bool ss(pair<long long, long long> a, pair<long long, long long> b) {
  return (a.second < b.second);
}
long long ans[N];
vector<long long> adj[N];
void solve() {
  int n;
  cin >> n;
  memset(ans, -1, sizeof(ans));
  for (long long i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    adj[v].push_back(i);
    adj[u].push_back(i);
  }
  long long cnt = 0;
  for (long long i = 0; i < n; i++) {
    if (adj[i].size() > 2) {
      ans[adj[i][0]] = 0;
      ans[adj[i][1]] = 1;
      ans[adj[i][2]] = 2;
      cnt = 3;
      break;
    }
  }
  for (long long i = 1; i < n; i++) {
    if (ans[i] == -1) {
      ans[i] = cnt;
      cnt++;
    }
  }
  for (long long i = 1; i < n; i++) cout << ans[i] << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  ;
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}
