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 PYTHON3):

Solution:
#                                               |
#   _` |  __ \    _` |   __|   _ \   __ \    _` |   _` |
#  (   |  |   |  (   |  (     (   |  |   |  (   |  (   |
# \__,_| _|  _| \__,_| \___| \___/  _|  _| \__,_| \__,_|

import sys

def read_line():
	return sys.stdin.readline()[:-1]
 
def read_int():
	return int(sys.stdin.readline())
 
def read_int_line():
	return [int(v) for v in sys.stdin.readline().split()]

n = read_int()
d = {}
adj = {}
for i in range(n-1):
	u,v =read_int_line()
	adj[(u,v)] = -1
	if u not in d:
		d[u] =1 
	else:
		d[u]+=1
	if v not in d:
		d[v] =1 
	else:
		d[v]+=1
		
vertex = -1
for i in adj:
	if vertex != -1:
		break
	if d[i[0]] >=3 :
		vertex = i[0]
	if d[i[1]] >=3:
		vertex = i[1]

if vertex == -1:
	for i in range(n-1):
		print(i)
else:
	c = 0
	for i in adj:
		if c<3:
			if vertex in i:
				adj[i] = c
				c+=1
	l = list(adj.values())
	v = [i for i in range(3,n)]
	i = 0
	j = 0
	while i<n-1 and j<n-1:
		if l[i] != -1:
			print(l[i])
		else:
			print(v[j])
			j+=1
		i+=1