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 long M = 500200;
int n, l, r, t, m, current;
vector<int> edge[M];
pair<int, int> ans[M];
void search(int p, int last) {
  for (int i = 0; i < edge[p].size(); i++) {
    if (last != edge[p][i]) {
      current++;
      ans[edge[p][i]].first = current;
    }
  }
  current++;
  ans[p].second = current;
  for (int i = edge[p].size() - 1; i >= 0; i--) {
    if (last != edge[p][i]) {
      search(edge[p][i], p);
    }
  }
}
int main() {
  {
    ios ::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  }
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> l >> r;
    edge[l].push_back(r);
    edge[r].push_back(l);
  }
  current = 1;
  ans[1].first = 1;
  search(1, 0);
  for (int i = 1; i <= n; i++) {
    cout << ans[i].first << " " << ans[i].second << endl;
  }
  return 0;
}