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

inp = [int(x) for x in os.read(0, 2**25).split()]; ii = 0
n = inp[ii]; ii += 1

coupl = [[] for _ in range(n)]
for _ in range(n - 1):
    u = inp[ii] - 1; ii += 1
    v = inp[ii] - 1; ii += 1
    coupl[u].append(v)
    coupl[v].append(u)

bfs = [0]
found = [0]*n
children = coupl
for node in bfs:
    found[node] = 1
    for nei in coupl[node]:
        bfs.append(nei)
        c = coupl[nei]
        i = 0
        while c[i] != node:
            i += 1
        c[i], c[-1] = c[-1], c[i]
        c.pop()

family = [1]*n
for node in reversed(bfs):
    for child in children[node]:
        family[node] += family[child]

out = inp
out.append(0)
out[1] = 2 * n - 1

mark = [0]*n
for node in bfs:
    l = mark[node]
    r = l + 2 * family[node] - 1
    for child in children[node]:
        mark[child] = l
        l += 2 * family[child] - 1
        r -= 1
        out[2 * child + 1] = r
    out[2 * node] = r - 1

os.write(1, ' '.join(str(x + 1) for x in out))