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>
#pragma GCC optimize("O3")
using namespace std;
const int inf = 1e9;
const int maxsz = 1e6;
vector<int> tree[maxsz];
bool visit[maxsz];
pair<int, int> ans[maxsz];
int sz = 0;
int dfs(int vert, int left, int right, int father) {
  visit[vert] = true;
  int sum = 0;
  int new_left = right;
  int new_right = right;
  for (int next : tree[vert]) {
    if (next == father) {
      continue;
    }
    new_left--;
    new_right++;
    new_right += tree[next].size() - 1;
    int cnt = dfs(next, new_left, new_right, vert);
    new_right += 2 * cnt - tree[next].size() - 1;
    sum += cnt;
  }
  ans[vert] = make_pair(left, right);
  return sum + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << setprecision(20);
  cout << fixed;
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int left, right;
    cin >> left >> right;
    tree[left - 1].push_back(right - 1);
    tree[right - 1].push_back(left - 1);
  }
  dfs(0, 1, tree[0].size() + 2, -1);
  for (int i = 0; i < n; i++) {
    cout << ans[i].first << " " << ans[i].second << "\n";
  }
  return 0;
}