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;
using ll = long long;
const int N = 5 * (int)1e5;
vector<int> conns[N];
pair<int, int> res[N];
int siz[N];
int dfs(int i, int p) {
  siz[i] = 1;
  for (auto t : conns[i]) {
    if (t != p) siz[i] += dfs(t, i);
  }
  return siz[i];
}
void solve(int i, int p, int a, int b) {
  int m = (int)conns[i].size() - (p != i);
  int x = b + m;
  res[i] = {a, x};
  int j = 1;
  int s = 1;
  for (auto t : conns[i]) {
    if (t == p) continue;
    solve(t, i, x - j, x + s);
    s += 2 * siz[t] - 1;
    ++j;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    conns[a].push_back(b);
    conns[b].push_back(a);
  }
  dfs(0, 0);
  solve(0, 0, 0, 1);
  for (int i = 0; i < n; ++i) {
    cout << res[i].first + 1 << ' ' << res[i].second + 1 << '\n';
  }
}