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 N = 1e5 + 9, MOD = 1e9 + 7;
int n, ans[N];
vector<pair<int, int>> G[N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    G[--u].push_back({--v, i});
    G[v].push_back({u, i});
  }
  int id = 0;
  memset(ans, -1, n * sizeof ans[0]);
  for (int i = 0; i < n && id < 3; ++i) {
    if (G[i].size() == 1 && ans[G[i][0].second] == -1)
      ans[G[i][0].second] = id++;
  }
  for (int i = 0; i < n - 1; ++i) {
    if (ans[i] == -1) ans[i] = id++;
    printf("%d\n", ans[i]);
  }
  return 0;
}