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;
vector<int> v[500100];
int f[500100];
int e[500100];
int mm[500100];
vector<int> dd;
void go(int cr, int p, int o) {
  if (v[cr].size() == 1 && p != 0) {
    e[cr] = o;
    mm[cr] = e[cr];
    return;
  }
  int last;
  if (p != 0) o--;
  last = o + v[cr].size() + 1;
  o += v[cr].size() - 1;
  e[cr] = o + 1;
  for (int i = v[cr].size() - 1; i >= 0; i--) {
    if (p == v[cr][i]) continue;
    f[v[cr][i]] = o;
    go(v[cr][i], cr, last);
    last = mm[v[cr][i]] + 1;
    mm[cr] = max(mm[cr], mm[v[cr][i]]);
    o--;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  f[1] = 1;
  go(1, 0, 2);
  for (int i = 1; i <= n; i++) printf("%d %d \n", f[i], e[i]);
  return 0;
}