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

Source: codeforces | Difficulty: HARD
Tags: Tree algorithms

Divide and conquer

Graph traversal

Constructive algorithms


Solutions:
def main():
	import sys
	input = sys.stdin.buffer.readline
	N = int(input())
	adj = [[] for _ in range(N + 1)]
	for _ in range(N - 1):
		(a, b) = map(int, input().split())
		adj[a].append(b)
		adj[b].append(a)
	st = [1]
	seen = [0] * (N + 1)
	seen[1] = 1
	LL = [0] * (2 * N + 3)
	LL[-1] = 1
	LL[1] = 0
	LL_rev = [0] * (2 * N + 3)
	LL_rev[1] = -1
	LL_rev[-1] = 0
	st_append = st.append
	while st:
		v = st.pop()
		for u in adj[v]:
			if not seen[u]:
				seen[u] = 1
				st_append(u)
				v_next = LL[v]
				v_prev = LL_rev[v]
				LL[v] = u
				LL[u] = v_next
				LL_rev[v_next] = u
				LL_rev[u] = v
				u_ = -u
				LL_rev[v] = u_
				LL[v_prev] = u_
				LL[u_] = v
	s = -1
	ans1 = [0] * (N + 1)
	ans2 = [0] * (N + 1)
	for i in range(1, 2 * N + 1):
		if s < 0:
			ans1[-s] = i
		else:
			ans2[s] = i
		s = LL[s]
	ans = '\n'.join([' '.join(map(str, [ans1[i], ans2[i]])) for i in range(1, N + 1)])
	sys.stdout.write(ans)
main()