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;
const int MAX_N = 5e5 + 5;
int available = 1;
vector<int> tree[MAX_N];
int left_label[MAX_N], right_label[MAX_N];
void dfs(int v, int parent_v) {
  int no_of_children = tree[v].size() - (parent_v == 0 ? 0 : 1);
  available += no_of_children + 1;
  right_label[v] = available;
  for (int i = 0, child_no = 0; i < tree[v].size(); i++) {
    int child_v = tree[v][i];
    if (child_v == parent_v) {
      continue;
    }
    child_no++;
    left_label[child_v] = right_label[v] - child_no;
    dfs(child_v, v);
  }
}
int main() {
  int no_of_vertices;
  scanf("%d", &no_of_vertices);
  int no_of_edges = no_of_vertices - 1;
  for (int i = 1; i <= no_of_edges; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    tree[u].push_back(v);
    tree[v].push_back(u);
  }
  left_label[1] = available;
  dfs(1, 0);
  for (int i = 1; i <= no_of_vertices; i++) {
    cout << left_label[i] << " " << right_label[i] << "\n";
  }
  return 0;
}