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 N = 6e5 + 6;
const long long MOD = 1e9 + 7;
pair<int, int> pat[N];
vector<int> g[N];
int dfs(int v, int ind, int par = -1) {
  int qan = g[v].size() - 1;
  if (par == -1) qan++;
  pat[v].second = ind + qan;
  int tv = pat[v].second;
  for (auto to : g[v]) {
    if (to == par) continue;
    pat[to].first = ind++;
  }
  reverse(g[v].begin(), g[v].end());
  for (auto to : g[v]) {
    if (to == par) continue;
    tv = dfs(to, tv + 1, v);
  }
  return tv;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cout.precision(13);
  cout.setf(ios::showpoint);
  int n;
  cin >> n;
  for (int i = 0; i < (n - 1); ++i) {
    int goa, gob;
    cin >> goa >> gob;
    g[goa].push_back(gob);
    g[gob].push_back(goa);
  }
  pat[1].first = 0;
  dfs(1, 1);
  for (int i = (1); i <= (n); ++i) {
    cout << pat[i].first + 1 << " " << pat[i].second + 1 << endl;
  }
  return 0;
}