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;
long long n, l[500010], r[500010], pont, jafoi[500010];
vector<long long> vertice[500010];
void dfs(long long x) {
  jafoi[x]++;
  long long pont_aux = pont + vertice[x].size() - 1 - 1;
  if (x == 1) pont_aux++;
  r[x] = pont_aux + 1;
  pont = r[x] + 1;
  for (long long i = 0; i < vertice[x].size(); i++) {
    long long u = vertice[x][i];
    if (jafoi[u] != 0) continue;
    l[u] = pont_aux;
    pont_aux--;
    dfs(u);
  }
}
void solve() {
  cin >> n;
  for (long long i = 1; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    vertice[a].push_back(b);
    vertice[b].push_back(a);
  }
  pont = 2;
  l[1] = 1;
  dfs(1);
  for (long long i = 1; i <= n; i++) cout << l[i] << " " << r[i] << "\n";
}
signed main() { solve(); }