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> e[555555];
int tsize[555555];
void dfs(int now, int p) {
  tsize[now] = 1;
  for (int k : e[now])
    if (k != p) dfs(k, now), tsize[now] += tsize[k];
}
pair<int, int> r[555555];
void solve(int now, int p, int l, int r) {
  int xx = r - 1, x = 0;
  for (int k : e[now])
    if (k != p) {
      x += e[k].size();
      ::r[k] = {xx, r + x};
      solve(k, now, xx, r + x);
      xx--, x += (tsize[k] - 1) * 2 - (e[k].size() - 1);
    }
}
int main() {
  int n, a, b;
  scanf("%d", &n);
  for (int i = 1; i < n; i++)
    scanf("%d%d", &a, &b), e[a].push_back(b), e[b].push_back(a);
  dfs(1, 1);
  r[1] = {1, e[1].size() + 2};
  solve(1, 1, 1, e[1].size() + 2);
  for (int i = 1; i < n + 1; i++) printf("%d %d\n", r[i].first, r[i].second);
}