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<vector<int>> neighbors;
vector<pair<int, int>> answers;
int k = 0;
void dfs(int i, int parent) {
  stack<int> order;
  for (auto& neighbor : neighbors[i])
    if (neighbor != parent) {
      answers[neighbor].first = ++k;
      order.push(neighbor);
    }
  answers[i].second = ++k;
  while (!order.empty()) {
    auto top = order.top();
    order.pop();
    dfs(top, i);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n;
  cin >> n;
  neighbors.resize(n);
  answers.resize(n);
  while (--n) {
    int x, y;
    cin >> x >> y;
    --x;
    --y;
    neighbors[x].push_back(y);
    neighbors[y].push_back(x);
  }
  answers[0].first = ++k;
  dfs(0, 0);
  for (auto& answer : answers)
    cout << answer.first << ' ' << answer.second << '\n';
}