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;
vector<int> a[500002];
int l[500002], r[500002], add[500002], dim[500002];
void dfs(int x, int tata) {
  int i, ll = a[x].size(), k = 0, p;
  dim[x] = 1;
  for (i = 0; i < ll; i++)
    if (a[x][i] != tata) {
      dfs(a[x][i], x);
      dim[x] += dim[a[x][i]];
      k++;
    }
  l[x] = 2 * dim[x] - k - 1;
  r[x] = 2 * dim[x];
  p = 2 * dim[x] - 1;
  k = 0;
  for (i = 0; i < ll; i++)
    if (a[x][i] != tata) {
      r[a[x][i]] = p;
      p -= 2 * dim[a[x][i]];
      add[a[x][i]] = k;
      k += 2 * dim[a[x][i]] - 1;
    }
}
void dfs2(int x, int tata) {
  int i, ll = a[x].size();
  l[x] += add[x];
  r[x] += add[x];
  for (i = 0; i < ll; i++)
    if (a[x][i] != tata) {
      add[a[x][i]] += add[x];
      dfs2(a[x][i], x);
    }
}
int main() {
  int n, i, x, y;
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    a[x].push_back(y);
    a[y].push_back(x);
  }
  dfs(1, 0);
  dfs2(1, 0);
  for (i = 1; i <= n; i++) cout << l[i] << " " << r[i] << '\n';
  return 0;
}