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 types import GeneratorType
input=sys.stdin.readline
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to

    return wrappedfunc
n=int(input())
class Graph:

    def __init__(self, V):
        self.V = V
        self.adj = [[] for i in range(V)]
    @bootstrap
    def DFSUtil(self, temp, v, visited,edges,parent,l,r):

        visited[v] = True

        temp.append(v)
        count=0
        for i in self.adj[v]:
            if visited[i] == False:
                temp = yield self.DFSUtil(temp, i, visited,edges,v,l,r)
                count+=1
        if count==0:

            edges[(v,parent)]=l[0]
            l[0]+=1
        else:
            edges[(v, parent)]=r[0]
            r[0]-=1
        yield temp

    def addEdge(self, v, w):
        self.adj[v].append(w)
        self.adj[w].append(v)

G=Graph(n)
all_edges=[]
for _ in range(n-1):
    u,v=list(map(int,input().split()))
    u-=1
    v-=1
    G.addEdge(u,v)
    all_edges.append((u,v))

l=[0]
r=[n-2]
visited=[False]*n
edges={}
G.DFSUtil([],0,visited,edges,-1,l,r)
for i in range(0,len(all_edges)):
    if all_edges[i] in edges:
        print(edges[all_edges[i]])
    else:
        print(edges[(all_edges[i][1],all_edges[i][0])])
