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;
int n, m[2 * 500007], curr, l[500007], r[500007];
vector<long long> a[500007];
void dfs(int v, int par) {
  for (auto to : a[v])
    if (to != par) m[++curr] = to;
  m[++curr] = v;
  for (int i = a[v].size() - 1; i >= 0; --i) {
    int to = a[v][i];
    if (to != par) dfs(to, v);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int v1, v2;
    cin >> v1 >> v2;
    a[v1].push_back(v2);
    a[v2].push_back(v1);
  }
  m[1] = 1;
  curr = 1;
  dfs(1, -1);
  for (int i = 1; i <= 2 * n; ++i) {
    if (l[m[i]] == 0) l[m[i]] = i;
    r[m[i]] = i;
  }
  for (int i = 1; i <= n; ++i) cout << l[i] << ' ' << r[i] << endl;
  return 0;
}