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 long double eps = 1e-9;
const long double pi = acos(-1.0);
const long long INF = (long long)(2e18);
const int inf = (int)(1e9 + 7);
vector<pair<int, int> > res;
vector<vector<int> > g;
int cnt = 1;
void dfs(int v, int p) {
  reverse((g[v]).begin(), (g[v]).end());
  for (auto& it : g[v]) {
    if (it == p) continue;
    res[it].first = cnt++;
  }
  res[v].second = cnt++;
  reverse((g[v]).begin(), (g[v]).end());
  for (auto& it : g[v]) {
    if (it == p) continue;
    dfs(it, v);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  cin.tie(nullptr);
  int n;
  cin >> n;
  g.assign(n, vector<int>());
  res.assign(n, pair<int, int>());
  for (int i = 0; i < n - 1; ++i) {
    int f, t;
    cin >> f >> t;
    --f;
    --t;
    g[f].push_back(t);
    g[t].push_back(f);
  }
  res[0].first = cnt++;
  dfs(0, 0);
  for (auto& it : res) cout << it.first << " " << it.second << endl;
  return 0;
}