Problem: We had a really tough time generating tests for problem D. In order to prepare strong tests, we had to solve the following problem.

Given an undirected labeled tree consisting of n vertices, find a set of segments such that:

  1. both endpoints of each segment are integers from 1 to 2n, and each integer from 1 to 2n should appear as an endpoint of exactly one segment; 
  2. all segments are non-degenerate; 
  3. for each pair (i, j) such that i ≠ j, i ∈ [1, n] and j ∈ [1, n], the vertices i and j are connected with an edge if and only if the segments i and j intersect, but neither segment i is fully contained in segment j, nor segment j is fully contained in segment i. 



Can you solve this problem too?

Input

The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of vertices in the tree.

Then n - 1 lines follow, each containing two integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) denoting the endpoints of the i-th edge.

It is guaranteed that the given graph is a tree.

Output

Print n pairs of integers, the i-th pair should contain two integers l_i and r_i (1 ≤ l_i < r_i ≤ 2n) — the endpoints of the i-th segment. All 2n integers you print should be unique.

It is guaranteed that the answer always exists.

Examples

Input


6
1 2
1 3
3 4
3 5
2 6


Output


9 12
7 10
3 11
1 5
2 4
6 8


Input


1


Output


1 2

Solution (in PYTHON):

Solution:
import sys
import os
range = xrange
input = raw_input
 
S = os.read(0, 2**25)
n = 0
i = 0
while S[i] >= '0':
    n = 10 * n + ord(S[i]) - 48
    i += 1
i += 2
 
coupl = [[] for _ in range(n)]
for _ in range(n - 1):
    u = v = 0
    while ord(S[i]) >= 48:
        u = 10 * u + ord(S[i]) - 48
        i += 1
    i += 1
    while ord(S[i]) >= 48:
        v = 10 * v + ord(S[i]) - 48
        i += 1
    i += 2
    coupl[u - 1].append(v - 1)
    coupl[v - 1].append(u - 1)
 
bfs = [0]
children = coupl
for node in bfs:
    for nei in coupl[node]:
        d = coupl[nei]
        i = 0
        while d[i] != node:
            i += 1
        d[i] = d[-1]
        d.pop()
        bfs.append(nei)
 
family = [2]*n
for node in reversed(bfs):
    for child in children[node]:
        family[node] += family[child]
 
out = [2 * n - 1]*(2 * n)
out[1] = 2 * n 
for node in bfs:
    r = out[2 * node]
    l = r - (family[node] - 1)
    for child in children[node]:
        out[2 * child] = l = l + family[child] - 1
        out[2 * child + 1] = r
        r -= 1
    out[2 * node] = r
 
import cStringIO
stream = cStringIO.StringIO()
f = stream.write
for x in out:
    f(str(x))
    f(' ')
 
os.write(1, stream.getvalue())