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;
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
const long long inf = 4e18;
const int infi = 2e9;
int i, j, n, m, k, qq = 1, fr, to, w, x, y, mod = 998244353;
char ch;
vector<vector<int> > g;
vector<int> ans;
int max_pos;
void dfs(int second, int pr, int pos) {
  ans[pos] = second;
  ans[max_pos + g[second].size() + (second == pr)] = second;
  max_pos += g[second].size() + (second == pr);
  int cur_pos = max_pos;
  int cnt = 0;
  for (int i = 0; i < g[second].size(); i++) {
    int to = g[second][i];
    if (to != pr) {
      cnt++;
      dfs(to, second, cur_pos - cnt);
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  g.resize(n);
  ans.resize(2 * n, -1);
  for (int i = 1; i < n; i++) {
    cin >> fr >> to;
    fr--;
    to--;
    g[fr].push_back(to);
    g[to].push_back(fr);
  }
  dfs(0, 0, 0);
  vector<vector<int> > v(n);
  for (int i = 0; i < 2 * n; i++) {
    v[ans[i]].push_back(i + 1);
  }
  for (auto x : v) {
    cout << x[0] << ' ' << x[1] << '\n';
  }
}