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<pair<long long, long long>> res(500009);
long long counter = 1;
void dfs_tree(long long idx, long long par, vector<vector<long long>> &adj) {
  for (auto i : adj[idx]) {
    if (i != par) {
      res[i].first = counter++;
    }
  }
  res[idx].second = counter++;
  reverse((adj[idx]).begin(), (adj[idx]).end());
  for (auto i : adj[idx]) {
    if (i != par) {
      dfs_tree(i, idx, adj);
    }
  }
}
int main() {
  long long(n);
  scanf("%lld", &(n));
  ;
  vector<vector<long long>> adj(n);
  for (long long i = 0; i < n - 1; i++) {
    long long a, b;
    scanf("%lld %lld", &a, &b);
    adj[a - 1].push_back(b - 1);
    adj[b - 1].push_back(a - 1);
  };
  res[0].first = counter++;
  dfs_tree(0, -1, adj);
  for (long long i = 0; i < (n); i++) {
    printf("%lld %lld\n", res[i].first, res[i].second);
  }
  return 0;
}