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>
import sys;
from sys import stdin, stdout
adj = [[] for i in range(10**5+7)];

#def dfs()


n = int(input());
query = []
for i in range(n-1):
    u,v = map(int,input().split());
    query.append([u,v]);
    adj[u].append(v);
    adj[v].append(u);

flag = True;
vertex = -1;
for i in range(n+1):
    if(len(adj[i])>=3):
        flag=False;
        vertex=i;
        
if(flag):
    for i in range(n-1):
#        print(i);
        stdout.write(str(i)+"\n")
    sys.exit();
color = [adj[vertex][0],adj[vertex][1],adj[vertex][2]];
cnt = 3;
for i in range(n-1):
    if(query[i][0]==vertex and query[i][1] in color):
        #print(color.index(query[i][1]));
        stdout.write(str(color.index(query[i][1]))+"\n")
    elif(query[i][1]==vertex and query[i][0] in color):
        #print(color.index(query[i][0]));
        stdout.write(str(color.index(query[i][0]))+"\n")
    else:
        stdout.write(str(cnt)+"\n");
        #print(cnt);
        cnt+=1;

        
