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;
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  os << "]";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, set<T> S) {
  os << "{ ";
  for (auto s : S) os << s << " ";
  return os << "}";
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <class L, class R>
ostream &operator<<(ostream &os, map<L, R> M) {
  os << "{ ";
  for (auto m : M) os << "(" << m.first << ":" << m.second << ") ";
  return os << "}";
}
vector<long long> adj[500050];
long long a[500050], b[500050];
set<long long> s;
void dfs(long long x, long long pr) {
  for (long long i = 0; i < adj[x].size(); i++)
    if (adj[x][i] != pr) a[adj[x][i]] = *s.begin(), s.erase(s.begin());
  b[x] = *s.begin();
  s.erase(s.begin());
  for (long long j = adj[x].size() - 1; j >= 0; j--)
    if (adj[x][j] != pr) dfs(adj[x][j], x);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, x, y;
  cin >> n;
  if (n == 1) {
    cout << "1 2";
    return 0;
  }
  for (long long i = 2; i < 2 * n + 1; i++) s.insert(i);
  for (long long i = 0; i < n - 1; i++)
    cin >> x >> y, adj[x].push_back(y), adj[y].push_back(x);
  a[1] = 1;
  dfs(1, -1);
  for (long long i = 1; i < n + 1; i++) cout << a[i] << " " << b[i] << "\n";
  return 0;
}