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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<int> adjlist[500000];
vector<int> children[500000];
pair<int, int> ans[500000];
int dfs(int v, int p, int s) {
  for (int i : adjlist[v]) {
    if (i != p) children[v].push_back(i);
  }
  ans[v].second = s + children[v].size();
  s = ans[v].second + 1;
  for (int i = 0; i < children[v].size(); ++i) {
    ans[children[v][i]].first = ans[v].second - i - 1;
    s = dfs(children[v][i], v, s) + 1;
  }
  return s - 1;
}
int main() {
  int n, x, y;
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    scanf("%d %d", &x, &y);
    adjlist[x - 1].push_back(y - 1);
    adjlist[y - 1].push_back(x - 1);
  }
  ans[0].first = 1;
  dfs(0, -1, 2);
  for (int i = 0; i < n; ++i) printf("%d %d\n", ans[i].first, ans[i].second);
  return 0;
}