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> adj[500105];
int xMove[] = {0, 1, 0, -1, 1, 1, -1, -1, 2, 2, -2, -2, 1, 1, -1, -1};
int yMove[] = {1, 0, -1, 0, 1, -1, 1, -1, 1, -1, 1, -1, 2, -2, 2, -2};
array<int, 2> segs[500105];
int cur = 2;
int visited[500105];
void dfs(int u) {
  visited[u] = 1;
  for (int v : adj[u]) {
    if (!visited[v]) segs[v][0] = cur++;
  }
  segs[u][1] = cur++;
  for (int j = adj[u].size() - 1; j >= 0; j--)
    if (!visited[adj[u][j]]) dfs(adj[u][j]);
}
int main() {
  cin.tie(0), cout.tie(0);
  ios_base::sync_with_stdio(0);
  cout.setf(ios::fixed);
  cout.precision(10);
  int TC = 1;
  int n, m, k, q;
  {
    cin >> n;
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      cin >> u >> v;
      if (u > v) swap(u, v);
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    segs[1][0] = 1;
    dfs(1);
    for (int i = 1; i <= n; i++)
      cout << segs[i][0] << " " << segs[i][1] << "\n";
  }
  return 0;
}