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;
vector<int> adj[100000];
int vis[100000];
map<pair<int, int>, int> edges;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    vis[i] = -1;
  }
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    edges[make_pair(a, b)] = i;
    edges[make_pair(b, a)] = i;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  int counter = 0;
  for (int i = 0; i < n; i++) {
    if (adj[i].size() >= 3) {
      vis[edges[make_pair(i, adj[i][0])]] = 0;
      vis[edges[make_pair(i, adj[i][1])]] = 1;
      vis[edges[make_pair(i, adj[i][2])]] = 2;
      edges[make_pair(i, adj[i][0])] = -1;
      edges[make_pair(adj[i][0], i)] = -1;
      edges[make_pair(i, adj[i][1])] = -1;
      edges[make_pair(adj[i][1], i)] = -1;
      edges[make_pair(i, adj[i][2])] = -1;
      edges[make_pair(adj[i][2], i)] = -1;
      counter = 3;
      break;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < adj[i].size(); j++) {
      if (edges[make_pair(i, adj[i][j])] != -1) {
        vis[edges[make_pair(i, adj[i][j])]] = counter;
        counter++;
        edges[make_pair(i, adj[i][j])] = -1;
        edges[make_pair(adj[i][j], i)] = -1;
      }
    }
  }
  for (int i = 0; i < n - 1; i++) {
    cout << vis[i] << endl;
  }
}
