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;
const long long nax = 5e5 + 5;
vector<long long> graph[nax];
long long t = 1;
bool vis[nax];
long long x[nax], y[nax];
void dfs(long long cur, long long p) {
  for (long long i = graph[cur].size() - 1; i >= 0; i--) {
    long long node = graph[cur][i];
    if (node != p) {
      x[node] = t++;
    }
  }
  y[cur] = t++;
  for (long long u : graph[cur]) {
    if (u != p) dfs(u, cur);
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n;
  cin >> n;
  if (n == 1) {
    cout << "1 2" << endl;
    return 0;
  }
  for (long long i = 1; i <= n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  for (long long i = 1; i <= n; i++) {
    sort(graph[i].begin(), graph[i].end());
  }
  x[1] = t++;
  dfs(1, -1);
  for (long long i = 1; i <= n; i++) {
    cout << x[i] << " " << y[i] << endl;
  }
  return 0;
}